PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Simple harmonic
motion
Question Bank - Set 2
Liberty University
Question 1
Question
A particle is undergoing simple harmonic motion with an amplitude of 4 cm and
a period of 3 seconds. If the particle passes through the equilibrium position at
time t= 0, determine the displacement of the particle 2 seconds after passing
through the equilibrium position.
Solution
Step 1: Convert the period to angular frequency. Given: Amplitude, A= 4 cm
Period, T= 3 seconds
The equation relating period (T) and angular frequency (ω) is:
ω=2π
T
Plugging in the values, we get:
ω=2π
3
Step 2: Determine the displacement at t= 2 seconds. The general equation
for the displacement (x) of a particle undergoing simple harmonic motion at
time tis:
x=Acos(ωt)
Given that at t= 0, the particle passes through the equilibrium position
(which means x= 0), we have:
x= 4 cos 2π
3t
Now, to find the displacement at t= 2 seconds:
x= 4 cos 2π
3×2
x= 4 cos 4π
3
x= 4 cos 4π−3π
3
x= 4 cos π
3
x= 4 ×1
2
x= 2 cm
Therefore, the displacement of the particle 2 seconds after passing through
the equilibrium position is 2 cm.
Question 2
Question
A particle undergoes simple harmonic motion with an amplitude of 10 cm and
a period of 2 s. If the particle is at the equilibrium position at t = 0, determine
the displacement (in cm) of the particle when its velocity is half of its maximum
value.
Solution
Let’s denote the displacement of the particle from the equilibrium position at
time tas x(t) and the maximum velocity of the particle as Vmax. We know that
for a particle in simple harmonic motion, the velocity v(t) is given by:
v(t) = dx
dt =±ωpA2−x(t)2,
where Ais the amplitude of the motion, ω=2π
Tis the angular frequency,
and the positive/negative sign indicates direction.
Step 1: Calculate the angular frequency.
Given that the period T= 2 s, we can find the angular frequency as:
ω=2π
T=2π
2=πrad/s.
Step 2: Find the maximum displacement from the equilibrium position.
At the maximum velocity, the magnitude of the velocity is Vmax =ω·A.
Since the velocity is half its maximum value, v(t) = 0.5·Vmax = 0.5·ω·A.
2
Thus, 0.5·π·10 = ±π√100 −x2.
Step 3: Solve for x.
5 = p100 −x2
25 = 100 −x2
x2= 75
x=±√75 = ±5√3 cm.
Therefore, when the velocity of the particle is half of its maximum value,
the displacement of the particle from the equilibrium position is 5√3 cm.
Question 3
Question
A particle of mass mis attached to a spring with spring constant k. Initially,
the particle is at equilibrium. At time t= 0, the particle is displaced a distance
Afrom equilibrium and given an initial velocity of v0in the negative direction.
Determine the amplitude of the resulting simple harmonic motion.
Solution
Step 1: The equation of motion for simple harmonic motion is given by
md2x
dt2=−kx
where x(t) is the displacement of the particle from equilibrium at time t.
Step 2: The general solution to the equation of motion is x(t) = Acos(ωt+ϕ),
where Ais the amplitude, ω=qk
m, and ϕis the phase angle.
Step 3: Given the initial conditions x(0) = Aand v(0) = −v0, we can find
the amplitude A: At t= 0: x(0) = Acos(ϕ) = A v(0) = −Aωsin(ϕ) = −v0
Step 4: Dividing the two equations, we get:
v(0)
x(0) =−Aωsin(ϕ)
A=−ωsin(ϕ) = −v0
A
Step 5: Solving for ϕ:
−ωsin(ϕ) = −v0
A
sin(ϕ) = v0
Aω
3
ϕ=sin−1v0
Aω
Step 6: Using the initial condition x(0) = A, we can solve for the amplitude:
A=Acos(ϕ) = Acos sin−1v0
Aω
Step 7: A common approach is to square the equation x(0) = Aand write
it in terms of sine using the Pythagorean identity:
A2=A2cos2sin−1v0
Aω
1 = cos2sin−1v0
Aω
sin2sin−1v0
Aω = 1
v0
Aω 2
= 1
Step 8: Solving for A:
v2
0
A2ω2= 1
A2ω2=v2
0
A=v0
ω
Step 9: Therefore, the amplitude of the resulting simple harmonic motion is
A=v0
√k
m
.
Question 4
Question
A particle of mass mis attached to a vertical spring with spring constant k.
The particle is initially stretched downward from its equilibrium position and
released with an initial velocity upward. Show that the equation of motion for
the particle is given by my′′ +ky = 0, where y(t) represents the displacement
of the particle from equilibrium position at time t.
Solution
Let’s denote y(t) as the displacement of the particle from the equilibrium posi-
tion at time t. We can start by writing the forces acting on the particle at time
t: 1. The force due to gravity acts downward with magnitude mg. 2. The force
due to the spring acts upward with magnitude ky(t).
The net force Facting on the particle can be calculated by the second law
of motion:
F=ma =md2y
dt2
4
The force equation can be written as:
mg −ky(t) = md2y
dt2
Now, simplifying the equation we get:
my′′ +ky =mg
Therefore, the equation of motion for the particle is my′′ +ky = 0.
Question 5
Question
A mass-spring system is undergoing simple harmonic motion with an amplitude
of 0.1 m and a frequency of 5 Hz. At time t= 0, the mass is at its maximum
displacement from the equilibrium position and moving in the positive direction.
Determine the position of the mass at t= 0.02 s.
Solution
Let’s first express the equation of motion for simple harmonic motion:
x(t) = Acos(ωt +ϕ)
where: - x(t) is the position of the mass at time t, - Ais the amplitude of the
motion, - ωis the angular frequency, - ϕis the phase angle.
Step 1: Find the angular frequency ω.Given that the frequency f= 5
Hz, we have:
f=ω
2π
ω= 2πf = 2π×5 = 10πrad/s
Step 2: Find the phase angle ϕ.At t= 0, the mass is at the maximum
displacement and moving in the positive direction. This implies ϕ= 0.
Step 3: Determine the position of the mass at t= 0.02 s. Substitute
A= 0.1 m, ω= 10πrad/s, ϕ= 0, and t= 0.02 s into the equation of motion:
x(0.02) = 0.1 cos(10π×0.02 + 0)
x(0.02) = 0.1 cos(0.2π)
x(0.02) = 0.1 cos π
5
x(0.02) = 0.1×0.5878
x(0.02) ≈0.05878 m
Therefore, the position of the mass at t= 0.02 s is approximately 0.05878
m.
5
Question 6
Question
A particle oscillates with simple harmonic motion along the x-axis with an
amplitude of 3 cm and a period of 2 seconds. If the particle is at x = 1 cm and
moving to the left at t = 0, find the equation of motion for the particle.
Solution
Step 1: Determine the angular frequency ωusing the formula:
ω=2π
T
where Tis the period of the motion.
ω=2π
2=π
Step 2: Since the particle is at x = 1 cm and moving to the left at t = 0,
the equation of motion can be written as:
x(t) = Asin(ωt +ϕ) + x0
where Ais the amplitude, ϕis the phase angle, and x0is the equilibrium position.
Step 3: Substitute the given values into the equation of motion: At t = 0, x
= 1 cm and the particle is moving to the left.
1 = 3 sin(ϕ)+0
Step 4: Solve for the phase angle ϕ:
sin(ϕ) = 1
3=⇒ϕ= sin−11
3≈19.47◦
Step 5: Substitute the values of A,ω,ϕ, and x0into the equation of motion:
x(t) = 3 sin(πt + 19.47◦)+0
Therefore, the equation of motion for the particle is:
x(t) = 3 sin(πt + 19.47◦)
Question 7
Question
A particle undergoes simple harmonic motion with an amplitude of 4 m and a
period of 2 seconds. If the particle is at its equilibrium position at time t= 0
and moving in the negative direction, find the displacement of the particle after
1 second.
6
Solution
Step 1: Find the angular frequency ωusing the formula ω=2π
T, where Tis the
period.
ω=2π
2=πrad/s
Step 2: Write the equation for simple harmonic motion as x(t) = Acos(ωt +ϕ),
where Ais the amplitude, ωis the angular frequency, and ϕis the phase angle.
Step 3: Since the particle is at its equilibrium position at time t= 0 and
moving in the negative direction, the equation becomes x(t) = −4 cos(πt).
Step 4: To find the displacement after 1 second, substitute t= 1 into the
equation.
x(1) = −4 cos(π·1) = −4 cos(π) = −4(−1) = 4 m
Therefore, the displacement of the particle after 1 second is 4 m.
Question 8
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and a
period of 2 seconds. If the particle pass through the equilibrium point at time
t= 0, find the displacement of the particle at t= 1.5 seconds.
Solution
Step 1: Find the angular frequency ω
The period Tis related to the angular frequency ωby the formula:
T=2π
ω
Given T= 2 seconds, plug this into the formula to find ω:
2 = 2π
ω
ω=2π
2=πrad/s
Step 2: Find the displacement at t= 1.5 seconds
The displacement xof the particle at time tis given by:
x(t) = Acos(ωt)
where Ais the amplitude (5 cm in this case). Substituting the values given:
x(1.5) = 5 cos(π×1.5) = 5 cos3π
2
x(1.5) = 5 ×0 = 0 cm
Therefore, the displacement of the particle at t= 1.5 seconds is 0 cm.
7
Question 9
Question
A particle of mass moscillates with simple harmonic motion along the x-axis
with an amplitude A. The maximum speed of the particle is vmax . Find the
maximum kinetic energy of the particle during its motion.
Solution
To find the maximum kinetic energy of the particle during its motion, we first
need to find the kinetic energy at the maximum speed, which occurs at the
equilibrium position where the particle is furthest from the equilibrium point.
Let’s denote the angular frequency of the oscillation as ω.
Step 1: Find the maximum kinetic energy at the equilibrium po-
sition At the equilibrium position, the kinetic energy Kof the particle can be
calculated as
K=1
2mv2
max
Step 2: Find the maximum kinetic energy To find the maximum
kinetic energy, we need to express vmax in terms of the amplitude A.
Given that the maximum speed vmax =Aω, we substitute this into the
equation found in Step 1:
K=1
2m(Aω)2
Step 3: Express ωin terms of AThe angular frequency ωcan be ex-
pressed in terms of Ausing the equation ω=2π
T, where Tis the period of the mo-
tion. Since the period Tcan be calculated as T=2π
ω, and the speed is given by
v=ωA, then the period of the motion Tcan be written as T=2π
ω=2π
v/A =2πA
v.
Step 4: Calculate ωin terms of ASubstitute ω=2π
T=2πv
2πA =v
Ainto
the expression for kinetic energy:
K=1
2m(A·v
A)2=1
2mv2
Hence, the maximum kinetic energy of the particle during its motion is
1
2mv2
max.
Question 10
Question
A mass-spring system oscillates with an amplitude of 0.2 meters and a period
of 2 seconds. If the mass is 0.5 kg, determine the maximum kinetic energy of
the mass during its motion.
8
Solution
Step 1: Find the angular frequency (ω) using the period (T).
ω=2π
T
ω=2π
2=πrad/s
Step 2: Calculate the maximum velocity of the mass using the amplitude
(A) and angular frequency (ω).
vmax =A·ω
vmax = 0.2·π= 0.2πm/s
Step 3: Determine the maximum kinetic energy (KEmax ) of the mass using
its maximum velocity.
KEmax =1
2mv2
max
KEmax =1
2·0.5·(0.2π)2
KEmax =1
2·0.5·(0.2)2·π2
KEmax = 0.05 ·0.04 ·π2
KEmax ≈0.0628 J
Therefore, the maximum kinetic energy of the mass during its motion is
approximately 0.0628 Joules.
Question 11
Question
A particle of mass 0.2 kg moves in simple harmonic motion with an amplitude of
0.1 m and a period of 2 seconds. Determine the maximum speed of the particle.
Solution
Step 1: First, we need to find the angular frequency of the motion. Given that
the period T= 2 seconds, we have the relation
T=2π
ω
where ωis the angular frequency. Solving for ω, we get
ω=2π
T=2π
2=πrad/s
9
Step 2: Next, we can find the equation for the velocity of the particle as
a function of time. The velocity of the particle in simple harmonic motion is
given by
v(t) = ωpA2−x(t)2
where Ais the amplitude of the motion and x(t) is the displacement of the
particle from the equilibrium position at time t. Thus, the maximum speed of
the particle occurs at x= 0. Substituting A= 0.1 m and ω=π, we get
vmax =π×√0.12=π×0.1 = 0.1πm/s
Therefore, the maximum speed of the particle is 0.1πm/s.
Question 12
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and
frequency of 2 Hz. If the displacement of the particle is 3 cm at time t= 0.1 s,
find the displacement of the particle at time t= 0.3 s.
Solution
1. We know the general equation for simple harmonic motion is given as:
x(t) = Acos(2πft +ϕ)
where: - A= amplitude of motion (5 cm in this case), - f= frequency of motion
(2 Hz in this case), - t= time, and - ϕ= phase angle.
2. Given that the displacement of the particle is 3 cm at t= 0.1 s:
3 = 5 cos(2π×2×0.1 + ϕ)
3. We can solve the above equation to find the phase angle, ϕ.
4. Now, to find the displacement at t= 0.3 s:
x(0.3) = 5 cos(2π×2×0.3 + ϕ)
5. Substitute the phase angle ϕobtained from step 3 into the above equation
and calculate the displacement at t= 0.3 s.
Question 13
Question
A mass-spring system undergoes simple harmonic motion with an amplitude of
0.2 m and a period of 2 s. If the maximum kinetic energy of the system is 2 J,
determine the total energy of the system.
10
Solution
Step 1: Identify the relevant formulas for the total energy of a mass-spring
system undergoing simple harmonic motion. The total energy (E) of the system
is the sum of the kinetic energy (KE) and potential energy (P E):
E=KE +P E
Step 2: Recall that for a mass-spring system, the kinetic energy at any point
is given by:
KE =1
2mω2(A2−x2)
where mis the mass, ωis the angular frequency, Ais the amplitude, and xis
the displacement from equilibrium.
Step 3: Since we are given the maximum kinetic energy, substitute x=A
into the kinetic energy equation:
KEmax =1
2mω2A2= 2 J
Step 4: Recall that the angular frequency (ω) is related to the period of the
motion as:
ω=2π
T
where Tis the period.
Step 5: Given that the period is 2 s, calculate the angular frequency:
ω=2π
2=πs−1
Step 6: Substituting ω=πs−1into the kinetic energy equation, we can solve
for the mass m:
2 = 1
2m(π)2(0.2)2
Step 7: Solve for mto find the mass of the system.
Step 8: Once you have found the mass, you can calculate the potential energy
using the relationship P E =1
2mω2x2. Substitute the values of m,ω, and A
into the potential energy equation.
Step 9: Finally, calculate the total energy Eof the system by summing up
the kinetic and potential energies.
Question 14
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and
a period of 2 seconds. If the particle is at a displacement of 3 cm from the
equilibrium position when the velocity is maximum, determine the equation of
motion of the particle.
11
Solution
Step 1: Find the angular frequency ω. Given: Amplitude, A= 5 cm, Period,
T= 2 s.
The angular frequency ωcan be found using the formula:
ω=2π
T
ω=2π
2=πrad/s
Step 2: Find the equation of motion. The general equation of simple har-
monic motion is:
x(t) = Acos(ωt +ϕ)
where: x(t) is the displacement at time t,Ais the amplitude, ωis the angular
frequency, ϕis the phase angle.
Given: A= 5 cm, ω=πrad/s, Maximum velocity occurs when the dis-
placement is 3 cm.
When the velocity is maximum, the displacement xis equal to the amplitude
A. So, x=A= 5 cm. This gives us the following equation:
5 = 5 cos(πt +ϕ)
Step 3: Find the phase angle ϕ. To find the phase angle ϕ, we substitute
t= 0 to the equation above:
5 = 5 cos(ϕ)⇒cos(ϕ) = 1 ⇒ϕ= 0
Step 4: Write the equation of motion. Substitute A= 5 cm, ω=π, and
ϕ= 0 into the general equation of motion:
x(t) = 5 cos(πt)
Therefore, the equation of motion of the particle is x(t) = 5 cos(πt).
Question 15
Question
A mass-spring system oscillates with an amplitude of 6 cm and a period of 2
seconds. If the maximum speed of the mass is 0.5 m/s, find the maximum
acceleration of the mass during its motion.
12
Solution
Step 1: Find the angular frequency (ω) of the system using the period (T).
Given: Amplitude, A= 6 cm = 0.06 m Period, T= 2 s
The relation between angular frequency and period is:
ω=2π
T
ω=2π
2=πrad/s
Step 2: Find the maximum acceleration (amax) of the mass using the angular
frequency and the amplitude. The equation for the acceleration of an object
undergoing simple harmonic motion is:
a=−ω2x
where xis the displacement from the equilibrium position.
At the amplitude, x=A:
amax =−ω2A
amax =−π2×0.06 m
amax =−0.115 m/s2
Therefore, the maximum acceleration of the mass during its motion is 0.115 m/s2.
Question 16
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and a
period of 2 seconds. If the particle is at its maximum displacement and moving
at a velocity of 10 cm/s, determine the displacement and velocity of the particle
after 1 second.
Solution
Step 1: Determine the angular frequency ωusing the period T.
ω=2π
T
Step 2: Calculate the angular frequency.
ω=2π
2=πrad/s
13
Step 3: Express the displacement xat t= 1 second using the amplitude and
angular frequency.
x(t) = Acos(ωt)
Step 4: Find the displacement at t= 1 second.
x(1) = 5 cos(π) = −5 cm
Step 5: Compute the velocity v(t) at t= 1 second.
v(t) = −Aω sin(ωt)
Step 6: Calculate the velocity at t= 1 second.
v(1) = −5πsin(π) = 0 cm/s
Therefore, after 1 second, the displacement of the particle is -5 cm, and its
velocity is 0 cm/s.
Question 17
Question
A particle of mass mis attached to a spring of spring constant k. The particle
is displaced a distance x0from equilibrium and released without initial velocity.
Find the total energy of the particle in terms of m,k, and x0.
Solution
To find the total energy of the particle, we need to consider both kinetic and
potential energy.
Step 1: Determining the potential energy The potential energy of a
spring is given by:
U=1
2kx2
where xis the displacement from equilibrium. Given that the particle is dis-
placed a distance x0from equilibrium, the potential energy when the particle is
at this position is:
U=1
2kx2
0
Step 2: Determining the kinetic energy The maximum displacement
x0occurs when the particle is at equilibrium, which is also where the maximum
speed is. This speed is given by v0=qk
mx0. Therefore, the kinetic energy
when the particle is at this position is:
K=1
2mv2
0=1
2m rk
mx0!2
=1
2kx2
0
14
Step 3: Finding the total energy The total energy Eof the particle is
the sum of its potential and kinetic energy:
E=U+K=1
2kx2
0+1
2kx2
0=kx2
0
Therefore, the total energy of the particle in terms of m,k, and x0is kx2
0.
Question 18
Question
A particle is executing simple harmonic motion with an angular frequency of
ω= 3 rad/s. If the maximum speed of the particle is 2 m/s, determine the
amplitude of the motion.
Solution
Step 1: The equation describing simple harmonic motion is x(t) = Acos(ωt +ϕ),
where Ais the amplitude, ωis the angular frequency, tis the time, and ϕis the
phase angle.
Step 2: The velocity of the particle is given by v(t) = −Aω sin(ωt +ϕ).
Step 3: The maximum speed occurs when the velocity is at its peak. In
other words, when sin(ωt +ϕ) = 1.
Step 4: Since the maximum speed is 2 m/s, we have 2 = Aω.
Step 5: Plugging in ω= 3 rad/s, we can solve for the amplitude A: 2 =
3A=⇒A=2
3m.
Therefore, the amplitude of the motion is 2
3meters.
Question 19
Question
A particle undergoes simple harmonic motion with an amplitude of 6 cm and a
period of 2 seconds. If its initial displacement is 4 cm and its initial velocity is
3 cm/s, determine the equation of motion for the particle.
Solution
Step 1: Determine the angular frequency ωusing the formula ω=2π
T, where T
is the period.
ω=2π
2=πrad/s
Step 2: The general equation of motion for simple harmonic motion is given
by x(t) = Acos(ωt) + Bsin(ωt).
15
Step 3: Substitute the given initial conditions into the equation to solve for
Aand B. At t= 0, x(0) = 4 and v(0) = 3.
x(0) = A= 4
Step 4: Differentiate the equation of motion to find the velocity function
v(t) = −Aω sin(ωt) + Bω cos(ωt).
Step 5: Substitute the initial velocity condition to solve for B.
v(0) = Bω = 3
B=3
π
Step 6: Substitute the values of Aand Bback into the general equation of
motion.
x(t) = 4 cos(πt) + 3
πsin(πt)
Therefore, the equation of motion for the particle is x(t) = 4 cos(πt) +
3
πsin(πt).
Question 20
Question
A particle undergoes simple harmonic motion with an amplitude of 8 cm and a
frequency of 2 Hz. If the particle is at its maximum displacement and moving
in the negative direction at time t= 0, what is the position function of the
particle in terms of time?
Solution
Step 1: Recall that the general formula for simple harmonic motion is given by
x(t) = Acos(2πft +ϕ), where Ais the amplitude, fis the frequency, and ϕis
the phase constant.
Step 2: Since the particle is at its maximum displacement and moving in
the negative direction at t= 0, the position function can be written as x(t) =
−8 cos(2π·2t+ϕ).
Step 3: To find the phase constant ϕ, we need to consider the initial con-
ditions. At t= 0, the particle is at its maximum displacement in the negative
direction. This means x(0) = −8.
Step 4: Substitute t= 0 and x(0) = −8 into the position function x(t):
−8 = −8 cos(0 + ϕ).
Step 5: Solve for ϕ: cos(ϕ) = 1 =⇒ϕ= 0.
Step 6: Therefore, the position function of the particle in terms of time is
x(t) = −8 cos(4πt).
16
Question 21
Question
A particle oscillates with simple harmonic motion according to the equation:
x(t) = 2 cos(3t+π/4)
Find the amplitude, period, frequency, and initial phase angle of the motion.
Solution
Given the equation for the simple harmonic motion:
x(t) = 2 cos(3t+π/4)
where x(t) represents the displacement of the particle at time t.
Step 1: Find the amplitude The amplitude of the motion is the coefficient
of the cosine function:
Amplitude = 2
Step 2: Find the period The period of the motion is given by the formula:
T=2π
ω
where ωis the angular frequency. From the equation x(t) = 2 cos(3t+π/4), we
have ω= 3. Thus, the period is:
T=2π
3
Step 3: Find the frequency The frequency of the motion is the reciprocal
of the period:
f=1
T=3
2π
Step 4: Find the initial phase angle The initial phase angle of the
motion can be determined from the equation:
Initial phase angle = π
4
Therefore, the amplitude is 2, the period is 2π
3, the frequency is 3
2π, and the
initial phase angle is π
4.
Question 22
Question
A block of mass mis attached to a spring with spring constant kand placed
on a horizontal frictionless surface. The block is pulled a distance Aaway from
the equilibrium position and released from rest. Calculate the maximum speed
of the block during its motion.
17
Solution
Let’s denote the displacement of the block from the equilibrium position as x(t)
and the maximum speed as vmax. We can use energy conservation to solve for
the maximum speed.
Step 1: The potential energy of the block-spring system at the maximum
displacement Ais equal to the initial kinetic energy of the block when released.
1
2kA2=1
2mv2
max
Step 2: Solve for the maximum speed vmax.
vmax =Ark
m
Therefore, the maximum speed of the block during its motion is vmax =
Aqk
m.
Question 23
Question
A particle of mass mis attached to a spring with force constant k. At time
t= 0, the particle is displaced from its equilibrium position by a distance A
and is given an initial velocity v0. Find the amplitude of the resulting simple
harmonic motion.
Solution
Step 1: Write down the equation of motion for simple harmonic motion. The
equation of motion for a particle undergoing simple harmonic motion is given
by:
md2x
dt2=−kx
Step 2: Solve the differential equation. We will assume the solution has the
form:
x(t) = Acos(ωt +ϕ)
where Ais the amplitude, ωis the angular frequency, and ϕis the phase angle.
Taking the first and second derivatives of x(t):
dx
dt =−Aωsin(ωt +ϕ)
d2x
dt2=−Aω2cos(ωt +ϕ)
18
Substitute x(t) and its derivatives into the equation of motion:
−mAω2cos(ωt +ϕ) = −kAcos(ωt +ϕ)
Step 3: Simplify the equation and find the amplitude. Comparing the coef-
ficients of cos(ωt +ϕ), we get:
mω2=k
ω=rk
m
Since x(0) = A, we have:
x(0) = A=Acos(ϕ)
ϕ= 0
So, the solution becomes:
x(t) = Acos(rk
mt)
Therefore, the amplitude of the resulting simple harmonic motion is equal
to the initial displacement, A.
Question 24
Question
A mass of 0.5 kg is attached to a spring with a spring constant of 40 N/m. The
mass is displaced 0.1 m from its equilibrium position and released. Determine
the amplitude, period, and angular frequency of the resulting simple harmonic
motion.
Solution
Step 1: We can find the amplitude of the simple harmonic motion using the
given displacement: Given displacement, x= 0.1 m. The amplitude of the
motion is the maximum displacement from the equilibrium position. Therefore,
the amplitude, A=|x|= 0.1 m.
Step 2: To find the period of the simple harmonic motion, we can use the
formula:
T=2π
ω
where Tis the period and ωis the angular frequency.
Step 3: We can find the angular frequency using the formula:
ω=rk
m
19
where kis the spring constant and mis the mass.
Step 4: Substitute the given values into the equations:
ω=r40
0.5
=√80
= 2√20
= 4√5 rad/s
Step 5: Substitute ωinto the formula for the period:
T=2π
4√5
=π
2√5s
Step 6: Therefore, the amplitude of the simple harmonic motion is 0.1 m,
the period is π
2√5seconds, and the angular frequency is 4√5 rad/s.
Question 25
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and
a period of 2 seconds. If the particle is at its maximum displacement at time
t= 0, find the displacement of the particle at time t= 1 second.
Solution
Step 1: Let’s first find the angular frequency, ω, of the simple harmonic motion.
We know that the formula for the period, T, of a particle undergoing simple
harmonic motion is given by:
T=2π
ω
Given that T= 2 seconds, we can solve for ω:
2 = 2π
ω
ω=2π
2=πrad/s
Step 2: Next, let’s express the displacement, x, of the particle at any time,
t, using the general formula for simple harmonic motion:
x(t) = Asin(ωt +ϕ)
20
where: A= amplitude = 5 cm, ω= angular frequency = πrad/s, and ϕ=
phase constant.
Step 3: Since the particle is at its maximum displacement at time t= 0, we
have:
x(0) = Asin(ϕ)=5
sin(ϕ)=1
ϕ=π
2
Step 4: Now, we can find the displacement of the particle at time t= 1
second:
x(1) = 5 sinπ·1 + π
2
x(1) = 5 sinπ+π
2
x(1) = 5 sin3π
2
x(1) = 5 ·(−1) = −5 cm
Therefore, the displacement of the particle at time t= 1 second is −5 cm.
Question 26
Question
An object of mass mis attached to a spring with spring constant k. The object
is displaced from its equilibrium position by a small distance x0and released
from rest. Determine the period Tof the resulting simple harmonic motion in
terms of m,k, and x0.
Solution
Step 1: Apply Hooke’s Law to find the force exerted by the spring at a displace-
ment xfrom equilibrium.
Step 1: F=−kx
Step 2: Apply Newton’s Second Law to the object to set up the differential
equation for simple harmonic motion.
Step 2: md2x
dt2=−kx
Step 3: Solve the differential equation by assuming x(t) = Acos(ωt), where
Ais the amplitude of the motion and ωis the angular frequency.
Step 3: d2x
dt2=−ω2Acos(ωt)
21
m(−ω2Acos(ωt)) = −kA cos(ωt)
m(−ω2) = −k
ω=rk
m
Step 4: The period Tof the motion is related to the angular frequency ωby
T=2π
ω.
Step 4: T=2π
ω=2π
qk
m
= 2πrm
k
Therefore, the period Tof the simple harmonic motion is 2πpm
k.
Question 27
Question
A mass-spring system oscillates with an amplitude of 0.2 m and a period of 2
seconds. If the mass is 0.5 kg, determine the equation describing the position
of the mass as a function of time.
Solution
Step 1: Find the angular frequency of the oscillation. Given that the period
T= 2 seconds, we can use the formula T=2π
ωto find the angular frequency ω.
2 = 2π
ω
ω=2π
2=πrad/s
Step 2: Determine the equation for the position of the mass. The equation
for simple harmonic motion is given by:
x(t) = Asin(ωt +ϕ)
where: A= amplitude = 0.2 m, ω= angular frequency = πrad/s.
To find the phase constant ϕ, we need to consider the initial conditions.
When t= 0, x= 0.2 m (amplitude).
x(0) = 0.2=0.2 sin(ϕ)
sin(ϕ)=1
Since sin(90) = 1, we have ϕ= 90 = π
2.
Therefore, the equation describing the position of the mass as a function of
time is:
x(t)=0.2 sinπt +π
2
22
Question 28
Question
A particle of mass mis attached to a spring with spring constant k. If the
particle is displaced from its equilibrium position by a distance of Aand released
from rest, find the speed of the particle when it is at a distance A
2from the
equilibrium position.
Solution
Step 1: Find the angular frequency ωof the simple harmonic motion.
Given: m, k, A
The angular frequency ωof the simple harmonic motion is given by:
ω=rk
m
Step 2: Find the amplitude of the motion at the instant the speed is required.
Given: A, A
2
The amplitude of the motion at the instant the speed is required is A
2.
Step 3: Find the displacement xof the particle from the equilibrium point
at the instant the speed is required. The displacement xis given by:
x=Acos(ωt)
At the instant the speed is required, x=A
2.
A
2=Acos(ωt) =⇒cos(ωt) = 1
2
Step 4: Find the velocity of the particle at the instant the speed is required.
The velocity of the particle is given by:
v=−Aω sin(ωt)
Substitute the values of Aand ωinto the equation:
v=−Ark
msin rk
mt!
Step 5: Find the speed of the particle when it is at a distance A
2from the
equilibrium position. Substitute x=A
2into the equation for velocity:
v=−A
2rk
m
23
v=−1
2ωA =−1
2rk
mA
So, the speed of the particle when it is at a distance A
2from the equilibrium
position is 1
2qk
mA.
Question 29
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and a
period of 2 seconds. If the particle starts at a displacement of 3 cm to the right
of the equilibrium position at time t= 0, find the displacement equation for the
particle.
Solution
Step 1: Identify the general form of the displacement equation for simple har-
monic motion. The general form of the displacement equation for simple har-
monic motion with an amplitude Aand period Tis given by:
x(t) = Acos 2π
Tt+ϕ
where x(t) is the displacement of the particle at time t,ϕis the phase angle, A
is the amplitude, and Tis the period.
Step 2: Find the values of amplitude Aand period T. Given that the
amplitude A= 5 cm and the period T= 2 seconds.
Step 3: Find the phase angle ϕ. To find the phase angle ϕ, substitute the
initial conditions into the general form of the displacement equation:
x(0) = 5 cos ϕ= 3
Solving for ϕ:
cos ϕ=3
5
ϕ= cos−13
5
Step 4: Substitute the values of A,T, and ϕinto the displacement equation.
Therefore, the displacement equation for the particle is:
x(t) = 5 cos 2π
2t+ cos−13
5
x(t) = 5 cosπt + cos−13
5
24
Question 30
Question
A particle moves in simple harmonic motion with a period of 5 seconds. If at t
= 0, the displacement of the particle is 0.3 m and its velocity is 0.1 m/s in the
positive direction, find an expression for the displacement of the particle as a
function of time.
Solution
Given: Period of motion, T = 5 seconds
Amplitude, A = 0.3 m
Initial displacement, x(0) = 0.3 m
Initial velocity, v(0) = 0.1 m/s
We know that the general equation for simple harmonic motion is given by:
x(t) = Acos(ωt +ϕ)
where: - A is the amplitude, - is the angular frequency (equal to 2π
T), - is the
phase angle.
To find , we use the formula ω=2π
Twhere T is the period. Substituting T
= 5 seconds, we get:
ω=2π
5
To find the phase angle , we use the initial conditions given:
x(0) = Acos(ϕ)=0.3
v(0) = −Aω sin(ϕ)=0.1
Solving these equations simultaneously, we find .
Step 1: Find the angular frequency
ω=2π
5
Step 2: Find the phase angle From the given initial conditions:
Acos(ϕ)=0.3
−Aω sin(ϕ)=0.1
Dividing the second equation by the first:
−ωtan(ϕ) = 0.1
0.3
tan(ϕ) = −0.1
0.3ω
25
Solving for , we get:
ϕ= arctan −0.1
0.3ω
Therefore, the displacement function for the particle as a function of time
is:
x(t)=0.3 cos 2π
5t+ arctan −0.1
0.3·2π
5
x(t) = 0.3 cos 2π
5t+ arctan −5
9π
Question 31
Question
A particle of mass mis attached to a spring with spring constant k. It is
displaced from its equilibrium position by a distance 2Aand released from rest.
What is the maximum speed of the particle during the subsequent motion?
Solution
Step 1: Find the angular frequency ωof the oscillation.
The angular frequency ωis given by:
ω=rk
m
Step 2: Find the amplitude of the oscillation.
The amplitude Ais half of the total displacement, so in this case A=
1
2(2A) = A.
Step 3: Find the maximum speed of the particle.
The maximum speed of the particle occurs when the displacement is the
amplitude, i.e., x=A.
The velocity of the particle at any position xis given by:
v=ωpA2−x2
At x=A:
vmax =ωpA2−A2
vmax =ω·0
vmax = 0
Therefore, the maximum speed of the particle during the subsequent motion
is 0.
26
Question 32
Question
A particle of mass mis attached to a spring with spring constant k. At time
t= 0, the particle is displaced Aunits from its equilibrium position x= 0
and released from rest. Find an expression for the velocity of the particle as a
function of time.
Solution
Step 1: First, we can write the equation of motion for simple harmonic motion.
The force acting on the particle is given by Hooke’s Law as F=−kx, where x
is the displacement from the equilibrium position. Using Newton’s Second Law,
F=ma, we have
−kx =md2x
dt2
Step 2: Rearranging the equation, we get
d2x
dt2+k
mx= 0
Step 3: The general solution to this second-order homogeneous differential
equation is given by
x(t) = Acos(ωt) + Bsin(ωt)
where ω=qk
m.
Step 4: We know that at time t= 0, the particle has been displaced Aunits
from the equilibrium position x= 0 and released from rest, i.e., x(0) = Aand
dx
dt t=0 = 0. Substituting these initial conditions into the general solution, we
find
x(t) = Acos(ωt)
Step 5: To find the velocity of the particle as a function of time, we differ-
entiate the position function with respect to time:
dx
dt =−Aωsin(ωt)
Step 6: Therefore, the velocity of the particle as a function of time is given
by
v(t) = −Aωsin(ωt)
27
Question 33
Question
A particle undergoes simple harmonic motion with an amplitude of 4 cm and a
period of 3 seconds. If at t= 0, the particle is at its maximum displacement,
find the equation of motion for the particle.
Solution
Given that the particle is at its maximum displacement (amplitude) at t= 0,
we can assume that the initial displacement x0= 4 cm and the initial velocity
v0= 0.
The general equation of motion for simple harmonic motion is given by:
x(t) = Acos(ωt +ϕ)
where: - Ais the amplitude, - ωis the angular frequency, - ϕis the phase angle.
From the given information, we know that A= 4 cm and T= 3 seconds.
The angular frequency ωcan be found using the formula T=2π
ω:
ω=2π
T=2π
3radians/second
Therefore, the equation of motion for the particle is:
x(t) = 4 cos 2π
3t+ϕ
To find the phase angle ϕ, we can use the initial conditions x(0) = Aand
v(0) = 0.
x(0) = 4 cos(ϕ)=4
cos(ϕ) = 1
ϕ= 0 radians
So, the final equation of motion for the particle is:
x(t) = 4 cos 2π
3t
Question 34
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and a
period of 4 seconds. If the particle is at its equilibrium position at t= 0 seconds
and moving in the positive direction, determine the displacement of the particle
at t= 1 second.
28
Solution
Step 1: Find the angular frequency ωusing the formula T=2π
ω, where Tis the
period. Step 2: Calculate the displacement at t= 1 second using the formula
x(t) = Acos(ωt), where x(t) is the displacement at time t. Step 3: Substitute
the values of A,ω, and tinto the formula to find the displacement.
Step 1: Given that the period T= 4 seconds, we use the formula
T=2π
ω
to find the angular frequency ω.
4 = 2π
ω=⇒ω=2π
4=π
2
Step 2: The displacement of the particle at any time tis given by
x(t) = Acos(ωt)
where Ais the amplitude. Given that the amplitude A= 5 cm, the equation
becomes
x(t) = 5 cos π
2t
Step 3: To find the displacement at t= 1 second, substitute t= 1 into the
equation.
x(1) = 5 cos π
2×1= 5 cos π
2= 5 ×0 = 0 cm
Therefore, the displacement of the particle at t= 1 second is 0 cm .
Question 35
Question
A block of mass mis attached to a spring with spring constant k. The block
is displaced from its equilibrium position and released. Determine the time for
the block to reach half of its amplitude for simple harmonic motion.
Solution
Let Abe the amplitude of the motion. The equation for the displacement xof
the block as a function of time tis given by x(t) = Asin(ωt), where ω=qk
m.
The time t1for the block to reach half of its amplitude can be obtained by
solving the equation x(t1) = A
2.
29
Now, to find the displacement at t= 2 seconds:
x= 4 cos 2π
3×2
x= 4 cos 4π
3
x= 4 cos 4π−3π
3
x= 4 cos π
3
x= 4 ×1
2
x= 2 cm
Therefore, the displacement of the particle 2 seconds after passing through
the equilibrium position is 2 cm.
Question 2
Question
A particle undergoes simple harmonic motion with an amplitude of 10 cm and
a period of 2 s. If the particle is at the equilibrium position at t = 0, determine
the displacement (in cm) of the particle when its velocity is half of its maximum
value.
Solution
Let’s denote the displacement of the particle from the equilibrium position at
time tas x(t) and the maximum velocity of the particle as Vmax. We know that
for a particle in simple harmonic motion, the velocity v(t) is given by:
v(t) = dx
dt =±ωpA2−x(t)2,
where Ais the amplitude of the motion, ω=2π
Tis the angular frequency,
and the positive/negative sign indicates direction.
Step 1: Calculate the angular frequency.
Given that the period T= 2 s, we can find the angular frequency as:
ω=2π
T=2π
2=πrad/s.
Step 2: Find the maximum displacement from the equilibrium position.
At the maximum velocity, the magnitude of the velocity is Vmax =ω·A.
Since the velocity is half its maximum value, v(t) = 0.5·Vmax = 0.5·ω·A.
2
Thus, 0.5·π·10 = ±π√100 −x2.
Step 3: Solve for x.
5 = p100 −x2
25 = 100 −x2
x2= 75
x=±√75 = ±5√3 cm.
Therefore, when the velocity of the particle is half of its maximum value,
the displacement of the particle from the equilibrium position is 5√3 cm.
Question 3
Question
A particle of mass mis attached to a spring with spring constant k. Initially,
the particle is at equilibrium. At time t= 0, the particle is displaced a distance
Afrom equilibrium and given an initial velocity of v0in the negative direction.
Determine the amplitude of the resulting simple harmonic motion.
Solution
Step 1: The equation of motion for simple harmonic motion is given by
md2x
dt2=−kx
where x(t) is the displacement of the particle from equilibrium at time t.
Step 2: The general solution to the equation of motion is x(t) = Acos(ωt+ϕ),
where Ais the amplitude, ω=qk
m, and ϕis the phase angle.
Step 3: Given the initial conditions x(0) = Aand v(0) = −v0, we can find
the amplitude A: At t= 0: x(0) = Acos(ϕ) = A v(0) = −Aωsin(ϕ) = −v0
Step 4: Dividing the two equations, we get:
v(0)
x(0) =−Aωsin(ϕ)
A=−ωsin(ϕ) = −v0
A
Step 5: Solving for ϕ:
−ωsin(ϕ) = −v0
A
sin(ϕ) = v0
Aω
3
ϕ=sin−1v0
Aω
Step 6: Using the initial condition x(0) = A, we can solve for the amplitude:
A=Acos(ϕ) = Acos sin−1v0
Aω
Step 7: A common approach is to square the equation x(0) = Aand write
it in terms of sine using the Pythagorean identity:
A2=A2cos2sin−1v0
Aω
1 = cos2sin−1v0
Aω
sin2sin−1v0
Aω = 1
v0
Aω 2
= 1
Step 8: Solving for A:
v2
0
A2ω2= 1
A2ω2=v2
0
A=v0
ω
Step 9: Therefore, the amplitude of the resulting simple harmonic motion is
A=v0
√k
m
.
Question 4
Question
A particle of mass mis attached to a vertical spring with spring constant k.
The particle is initially stretched downward from its equilibrium position and
released with an initial velocity upward. Show that the equation of motion for
the particle is given by my′′ +ky = 0, where y(t) represents the displacement
of the particle from equilibrium position at time t.
Solution
Let’s denote y(t) as the displacement of the particle from the equilibrium posi-
tion at time t. We can start by writing the forces acting on the particle at time
t: 1. The force due to gravity acts downward with magnitude mg. 2. The force
due to the spring acts upward with magnitude ky(t).
The net force Facting on the particle can be calculated by the second law
of motion:
F=ma =md2y
dt2
4
The force equation can be written as:
mg −ky(t) = md2y
dt2
Now, simplifying the equation we get:
my′′ +ky =mg
Therefore, the equation of motion for the particle is my′′ +ky = 0.
Question 5
Question
A mass-spring system is undergoing simple harmonic motion with an amplitude
of 0.1 m and a frequency of 5 Hz. At time t= 0, the mass is at its maximum
displacement from the equilibrium position and moving in the positive direction.
Determine the position of the mass at t= 0.02 s.
Solution
Let’s first express the equation of motion for simple harmonic motion:
x(t) = Acos(ωt +ϕ)
where: - x(t) is the position of the mass at time t, - Ais the amplitude of the
motion, - ωis the angular frequency, - ϕis the phase angle.
Step 1: Find the angular frequency ω.Given that the frequency f= 5
Hz, we have:
f=ω
2π
ω= 2πf = 2π×5 = 10πrad/s
Step 2: Find the phase angle ϕ.At t= 0, the mass is at the maximum
displacement and moving in the positive direction. This implies ϕ= 0.
Step 3: Determine the position of the mass at t= 0.02 s. Substitute
A= 0.1 m, ω= 10πrad/s, ϕ= 0, and t= 0.02 s into the equation of motion:
x(0.02) = 0.1 cos(10π×0.02 + 0)
x(0.02) = 0.1 cos(0.2π)
x(0.02) = 0.1 cos π
5
x(0.02) = 0.1×0.5878
x(0.02) ≈0.05878 m
Therefore, the position of the mass at t= 0.02 s is approximately 0.05878
m.
5
Question 6
Question
A particle oscillates with simple harmonic motion along the x-axis with an
amplitude of 3 cm and a period of 2 seconds. If the particle is at x = 1 cm and
moving to the left at t = 0, find the equation of motion for the particle.
Solution
Step 1: Determine the angular frequency ωusing the formula:
ω=2π
T
where Tis the period of the motion.
ω=2π
2=π
Step 2: Since the particle is at x = 1 cm and moving to the left at t = 0,
the equation of motion can be written as:
x(t) = Asin(ωt +ϕ) + x0
where Ais the amplitude, ϕis the phase angle, and x0is the equilibrium position.
Step 3: Substitute the given values into the equation of motion: At t = 0, x
= 1 cm and the particle is moving to the left.
1 = 3 sin(ϕ)+0
Step 4: Solve for the phase angle ϕ:
sin(ϕ) = 1
3=⇒ϕ= sin−11
3≈19.47◦
Step 5: Substitute the values of A,ω,ϕ, and x0into the equation of motion:
x(t) = 3 sin(πt + 19.47◦)+0
Therefore, the equation of motion for the particle is:
x(t) = 3 sin(πt + 19.47◦)
Question 7
Question
A particle undergoes simple harmonic motion with an amplitude of 4 m and a
period of 2 seconds. If the particle is at its equilibrium position at time t= 0
and moving in the negative direction, find the displacement of the particle after
1 second.
6
Solution
Step 1: Find the angular frequency ωusing the formula ω=2π
T, where Tis the
period.
ω=2π
2=πrad/s
Step 2: Write the equation for simple harmonic motion as x(t) = Acos(ωt +ϕ),
where Ais the amplitude, ωis the angular frequency, and ϕis the phase angle.
Step 3: Since the particle is at its equilibrium position at time t= 0 and
moving in the negative direction, the equation becomes x(t) = −4 cos(πt).
Step 4: To find the displacement after 1 second, substitute t= 1 into the
equation.
x(1) = −4 cos(π·1) = −4 cos(π) = −4(−1) = 4 m
Therefore, the displacement of the particle after 1 second is 4 m.
Question 8
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and a
period of 2 seconds. If the particle pass through the equilibrium point at time
t= 0, find the displacement of the particle at t= 1.5 seconds.
Solution
Step 1: Find the angular frequency ω
The period Tis related to the angular frequency ωby the formula:
T=2π
ω
Given T= 2 seconds, plug this into the formula to find ω:
2 = 2π
ω
ω=2π
2=πrad/s
Step 2: Find the displacement at t= 1.5 seconds
The displacement xof the particle at time tis given by:
x(t) = Acos(ωt)
where Ais the amplitude (5 cm in this case). Substituting the values given:
x(1.5) = 5 cos(π×1.5) = 5 cos3π
2
x(1.5) = 5 ×0 = 0 cm
Therefore, the displacement of the particle at t= 1.5 seconds is 0 cm.
7
Question 9
Question
A particle of mass moscillates with simple harmonic motion along the x-axis
with an amplitude A. The maximum speed of the particle is vmax . Find the
maximum kinetic energy of the particle during its motion.
Solution
To find the maximum kinetic energy of the particle during its motion, we first
need to find the kinetic energy at the maximum speed, which occurs at the
equilibrium position where the particle is furthest from the equilibrium point.
Let’s denote the angular frequency of the oscillation as ω.
Step 1: Find the maximum kinetic energy at the equilibrium po-
sition At the equilibrium position, the kinetic energy Kof the particle can be
calculated as
K=1
2mv2
max
Step 2: Find the maximum kinetic energy To find the maximum
kinetic energy, we need to express vmax in terms of the amplitude A.
Given that the maximum speed vmax =Aω, we substitute this into the
equation found in Step 1:
K=1
2m(Aω)2
Step 3: Express ωin terms of AThe angular frequency ωcan be ex-
pressed in terms of Ausing the equation ω=2π
T, where Tis the period of the mo-
tion. Since the period Tcan be calculated as T=2π
ω, and the speed is given by
v=ωA, then the period of the motion Tcan be written as T=2π
ω=2π
v/A =2πA
v.
Step 4: Calculate ωin terms of ASubstitute ω=2π
T=2πv
2πA =v
Ainto
the expression for kinetic energy:
K=1
2m(A·v
A)2=1
2mv2
Hence, the maximum kinetic energy of the particle during its motion is
1
2mv2
max.
Question 10
Question
A mass-spring system oscillates with an amplitude of 0.2 meters and a period
of 2 seconds. If the mass is 0.5 kg, determine the maximum kinetic energy of
the mass during its motion.
8
Solution
Step 1: Find the angular frequency (ω) using the period (T).
ω=2π
T
ω=2π
2=πrad/s
Step 2: Calculate the maximum velocity of the mass using the amplitude
(A) and angular frequency (ω).
vmax =A·ω
vmax = 0.2·π= 0.2πm/s
Step 3: Determine the maximum kinetic energy (KEmax ) of the mass using
its maximum velocity.
KEmax =1
2mv2
max
KEmax =1
2·0.5·(0.2π)2
KEmax =1
2·0.5·(0.2)2·π2
KEmax = 0.05 ·0.04 ·π2
KEmax ≈0.0628 J
Therefore, the maximum kinetic energy of the mass during its motion is
approximately 0.0628 Joules.
Question 11
Question
A particle of mass 0.2 kg moves in simple harmonic motion with an amplitude of
0.1 m and a period of 2 seconds. Determine the maximum speed of the particle.
Solution
Step 1: First, we need to find the angular frequency of the motion. Given that
the period T= 2 seconds, we have the relation
T=2π
ω
where ωis the angular frequency. Solving for ω, we get
ω=2π
T=2π
2=πrad/s
9
Step 2: Next, we can find the equation for the velocity of the particle as
a function of time. The velocity of the particle in simple harmonic motion is
given by
v(t) = ωpA2−x(t)2
where Ais the amplitude of the motion and x(t) is the displacement of the
particle from the equilibrium position at time t. Thus, the maximum speed of
the particle occurs at x= 0. Substituting A= 0.1 m and ω=π, we get
vmax =π×√0.12=π×0.1 = 0.1πm/s
Therefore, the maximum speed of the particle is 0.1πm/s.
Question 12
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and
frequency of 2 Hz. If the displacement of the particle is 3 cm at time t= 0.1 s,
find the displacement of the particle at time t= 0.3 s.
Solution
1. We know the general equation for simple harmonic motion is given as:
x(t) = Acos(2πft +ϕ)
where: - A= amplitude of motion (5 cm in this case), - f= frequency of motion
(2 Hz in this case), - t= time, and - ϕ= phase angle.
2. Given that the displacement of the particle is 3 cm at t= 0.1 s:
3 = 5 cos(2π×2×0.1 + ϕ)
3. We can solve the above equation to find the phase angle, ϕ.
4. Now, to find the displacement at t= 0.3 s:
x(0.3) = 5 cos(2π×2×0.3 + ϕ)
5. Substitute the phase angle ϕobtained from step 3 into the above equation
and calculate the displacement at t= 0.3 s.
Question 13
Question
A mass-spring system undergoes simple harmonic motion with an amplitude of
0.2 m and a period of 2 s. If the maximum kinetic energy of the system is 2 J,
determine the total energy of the system.
10
Solution
Step 1: Identify the relevant formulas for the total energy of a mass-spring
system undergoing simple harmonic motion. The total energy (E) of the system
is the sum of the kinetic energy (KE) and potential energy (P E):
E=KE +P E
Step 2: Recall that for a mass-spring system, the kinetic energy at any point
is given by:
KE =1
2mω2(A2−x2)
where mis the mass, ωis the angular frequency, Ais the amplitude, and xis
the displacement from equilibrium.
Step 3: Since we are given the maximum kinetic energy, substitute x=A
into the kinetic energy equation:
KEmax =1
2mω2A2= 2 J
Step 4: Recall that the angular frequency (ω) is related to the period of the
motion as:
ω=2π
T
where Tis the period.
Step 5: Given that the period is 2 s, calculate the angular frequency:
ω=2π
2=πs−1
Step 6: Substituting ω=πs−1into the kinetic energy equation, we can solve
for the mass m:
2 = 1
2m(π)2(0.2)2
Step 7: Solve for mto find the mass of the system.
Step 8: Once you have found the mass, you can calculate the potential energy
using the relationship P E =1
2mω2x2. Substitute the values of m,ω, and A
into the potential energy equation.
Step 9: Finally, calculate the total energy Eof the system by summing up
the kinetic and potential energies.
Question 14
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and
a period of 2 seconds. If the particle is at a displacement of 3 cm from the
equilibrium position when the velocity is maximum, determine the equation of
motion of the particle.
11
Solution
Step 1: Find the angular frequency ω. Given: Amplitude, A= 5 cm, Period,
T= 2 s.
The angular frequency ωcan be found using the formula:
ω=2π
T
ω=2π
2=πrad/s
Step 2: Find the equation of motion. The general equation of simple har-
monic motion is:
x(t) = Acos(ωt +ϕ)
where: x(t) is the displacement at time t,Ais the amplitude, ωis the angular
frequency, ϕis the phase angle.
Given: A= 5 cm, ω=πrad/s, Maximum velocity occurs when the dis-
placement is 3 cm.
When the velocity is maximum, the displacement xis equal to the amplitude
A. So, x=A= 5 cm. This gives us the following equation:
5 = 5 cos(πt +ϕ)
Step 3: Find the phase angle ϕ. To find the phase angle ϕ, we substitute
t= 0 to the equation above:
5 = 5 cos(ϕ)⇒cos(ϕ) = 1 ⇒ϕ= 0
Step 4: Write the equation of motion. Substitute A= 5 cm, ω=π, and
ϕ= 0 into the general equation of motion:
x(t) = 5 cos(πt)
Therefore, the equation of motion of the particle is x(t) = 5 cos(πt).
Question 15
Question
A mass-spring system oscillates with an amplitude of 6 cm and a period of 2
seconds. If the maximum speed of the mass is 0.5 m/s, find the maximum
acceleration of the mass during its motion.
12
Solution
Step 1: Find the angular frequency (ω) of the system using the period (T).
Given: Amplitude, A= 6 cm = 0.06 m Period, T= 2 s
The relation between angular frequency and period is:
ω=2π
T
ω=2π
2=πrad/s
Step 2: Find the maximum acceleration (amax) of the mass using the angular
frequency and the amplitude. The equation for the acceleration of an object
undergoing simple harmonic motion is:
a=−ω2x
where xis the displacement from the equilibrium position.
At the amplitude, x=A:
amax =−ω2A
amax =−π2×0.06 m
amax =−0.115 m/s2
Therefore, the maximum acceleration of the mass during its motion is 0.115 m/s2.
Question 16
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and a
period of 2 seconds. If the particle is at its maximum displacement and moving
at a velocity of 10 cm/s, determine the displacement and velocity of the particle
after 1 second.
Solution
Step 1: Determine the angular frequency ωusing the period T.
ω=2π
T
Step 2: Calculate the angular frequency.
ω=2π
2=πrad/s
13
Step 3: Express the displacement xat t= 1 second using the amplitude and
angular frequency.
x(t) = Acos(ωt)
Step 4: Find the displacement at t= 1 second.
x(1) = 5 cos(π) = −5 cm
Step 5: Compute the velocity v(t) at t= 1 second.
v(t) = −Aω sin(ωt)
Step 6: Calculate the velocity at t= 1 second.
v(1) = −5πsin(π) = 0 cm/s
Therefore, after 1 second, the displacement of the particle is -5 cm, and its
velocity is 0 cm/s.
Question 17
Question
A particle of mass mis attached to a spring of spring constant k. The particle
is displaced a distance x0from equilibrium and released without initial velocity.
Find the total energy of the particle in terms of m,k, and x0.
Solution
To find the total energy of the particle, we need to consider both kinetic and
potential energy.
Step 1: Determining the potential energy The potential energy of a
spring is given by:
U=1
2kx2
where xis the displacement from equilibrium. Given that the particle is dis-
placed a distance x0from equilibrium, the potential energy when the particle is
at this position is:
U=1
2kx2
0
Step 2: Determining the kinetic energy The maximum displacement
x0occurs when the particle is at equilibrium, which is also where the maximum
speed is. This speed is given by v0=qk
mx0. Therefore, the kinetic energy
when the particle is at this position is:
K=1
2mv2
0=1
2m rk
mx0!2
=1
2kx2
0
14
Step 3: Finding the total energy The total energy Eof the particle is
the sum of its potential and kinetic energy:
E=U+K=1
2kx2
0+1
2kx2
0=kx2
0
Therefore, the total energy of the particle in terms of m,k, and x0is kx2
0.
Question 18
Question
A particle is executing simple harmonic motion with an angular frequency of
ω= 3 rad/s. If the maximum speed of the particle is 2 m/s, determine the
amplitude of the motion.
Solution
Step 1: The equation describing simple harmonic motion is x(t) = Acos(ωt +ϕ),
where Ais the amplitude, ωis the angular frequency, tis the time, and ϕis the
phase angle.
Step 2: The velocity of the particle is given by v(t) = −Aω sin(ωt +ϕ).
Step 3: The maximum speed occurs when the velocity is at its peak. In
other words, when sin(ωt +ϕ) = 1.
Step 4: Since the maximum speed is 2 m/s, we have 2 = Aω.
Step 5: Plugging in ω= 3 rad/s, we can solve for the amplitude A: 2 =
3A=⇒A=2
3m.
Therefore, the amplitude of the motion is 2
3meters.
Question 19
Question
A particle undergoes simple harmonic motion with an amplitude of 6 cm and a
period of 2 seconds. If its initial displacement is 4 cm and its initial velocity is
3 cm/s, determine the equation of motion for the particle.
Solution
Step 1: Determine the angular frequency ωusing the formula ω=2π
T, where T
is the period.
ω=2π
2=πrad/s
Step 2: The general equation of motion for simple harmonic motion is given
by x(t) = Acos(ωt) + Bsin(ωt).
15
Step 3: Substitute the given initial conditions into the equation to solve for
Aand B. At t= 0, x(0) = 4 and v(0) = 3.
x(0) = A= 4
Step 4: Differentiate the equation of motion to find the velocity function
v(t) = −Aω sin(ωt) + Bω cos(ωt).
Step 5: Substitute the initial velocity condition to solve for B.
v(0) = Bω = 3
B=3
π
Step 6: Substitute the values of Aand Bback into the general equation of
motion.
x(t) = 4 cos(πt) + 3
πsin(πt)
Therefore, the equation of motion for the particle is x(t) = 4 cos(πt) +
3
πsin(πt).
Question 20
Question
A particle undergoes simple harmonic motion with an amplitude of 8 cm and a
frequency of 2 Hz. If the particle is at its maximum displacement and moving
in the negative direction at time t= 0, what is the position function of the
particle in terms of time?
Solution
Step 1: Recall that the general formula for simple harmonic motion is given by
x(t) = Acos(2πft +ϕ), where Ais the amplitude, fis the frequency, and ϕis
the phase constant.
Step 2: Since the particle is at its maximum displacement and moving in
the negative direction at t= 0, the position function can be written as x(t) =
−8 cos(2π·2t+ϕ).
Step 3: To find the phase constant ϕ, we need to consider the initial con-
ditions. At t= 0, the particle is at its maximum displacement in the negative
direction. This means x(0) = −8.
Step 4: Substitute t= 0 and x(0) = −8 into the position function x(t):
−8 = −8 cos(0 + ϕ).
Step 5: Solve for ϕ: cos(ϕ) = 1 =⇒ϕ= 0.
Step 6: Therefore, the position function of the particle in terms of time is
x(t) = −8 cos(4πt).
16
Question 21
Question
A particle oscillates with simple harmonic motion according to the equation:
x(t) = 2 cos(3t+π/4)
Find the amplitude, period, frequency, and initial phase angle of the motion.
Solution
Given the equation for the simple harmonic motion:
x(t) = 2 cos(3t+π/4)
where x(t) represents the displacement of the particle at time t.
Step 1: Find the amplitude The amplitude of the motion is the coefficient
of the cosine function:
Amplitude = 2
Step 2: Find the period The period of the motion is given by the formula:
T=2π
ω
where ωis the angular frequency. From the equation x(t) = 2 cos(3t+π/4), we
have ω= 3. Thus, the period is:
T=2π
3
Step 3: Find the frequency The frequency of the motion is the reciprocal
of the period:
f=1
T=3
2π
Step 4: Find the initial phase angle The initial phase angle of the
motion can be determined from the equation:
Initial phase angle = π
4
Therefore, the amplitude is 2, the period is 2π
3, the frequency is 3
2π, and the
initial phase angle is π
4.
Question 22
Question
A block of mass mis attached to a spring with spring constant kand placed
on a horizontal frictionless surface. The block is pulled a distance Aaway from
the equilibrium position and released from rest. Calculate the maximum speed
of the block during its motion.
17
Solution
Let’s denote the displacement of the block from the equilibrium position as x(t)
and the maximum speed as vmax. We can use energy conservation to solve for
the maximum speed.
Step 1: The potential energy of the block-spring system at the maximum
displacement Ais equal to the initial kinetic energy of the block when released.
1
2kA2=1
2mv2
max
Step 2: Solve for the maximum speed vmax.
vmax =Ark
m
Therefore, the maximum speed of the block during its motion is vmax =
Aqk
m.
Question 23
Question
A particle of mass mis attached to a spring with force constant k. At time
t= 0, the particle is displaced from its equilibrium position by a distance A
and is given an initial velocity v0. Find the amplitude of the resulting simple
harmonic motion.
Solution
Step 1: Write down the equation of motion for simple harmonic motion. The
equation of motion for a particle undergoing simple harmonic motion is given
by:
md2x
dt2=−kx
Step 2: Solve the differential equation. We will assume the solution has the
form:
x(t) = Acos(ωt +ϕ)
where Ais the amplitude, ωis the angular frequency, and ϕis the phase angle.
Taking the first and second derivatives of x(t):
dx
dt =−Aωsin(ωt +ϕ)
d2x
dt2=−Aω2cos(ωt +ϕ)
18
Substitute x(t) and its derivatives into the equation of motion:
−mAω2cos(ωt +ϕ) = −kAcos(ωt +ϕ)
Step 3: Simplify the equation and find the amplitude. Comparing the coef-
ficients of cos(ωt +ϕ), we get:
mω2=k
ω=rk
m
Since x(0) = A, we have:
x(0) = A=Acos(ϕ)
ϕ= 0
So, the solution becomes:
x(t) = Acos(rk
mt)
Therefore, the amplitude of the resulting simple harmonic motion is equal
to the initial displacement, A.
Question 24
Question
A mass of 0.5 kg is attached to a spring with a spring constant of 40 N/m. The
mass is displaced 0.1 m from its equilibrium position and released. Determine
the amplitude, period, and angular frequency of the resulting simple harmonic
motion.
Solution
Step 1: We can find the amplitude of the simple harmonic motion using the
given displacement: Given displacement, x= 0.1 m. The amplitude of the
motion is the maximum displacement from the equilibrium position. Therefore,
the amplitude, A=|x|= 0.1 m.
Step 2: To find the period of the simple harmonic motion, we can use the
formula:
T=2π
ω
where Tis the period and ωis the angular frequency.
Step 3: We can find the angular frequency using the formula:
ω=rk
m
19
where kis the spring constant and mis the mass.
Step 4: Substitute the given values into the equations:
ω=r40
0.5
=√80
= 2√20
= 4√5 rad/s
Step 5: Substitute ωinto the formula for the period:
T=2π
4√5
=π
2√5s
Step 6: Therefore, the amplitude of the simple harmonic motion is 0.1 m,
the period is π
2√5seconds, and the angular frequency is 4√5 rad/s.
Question 25
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and
a period of 2 seconds. If the particle is at its maximum displacement at time
t= 0, find the displacement of the particle at time t= 1 second.
Solution
Step 1: Let’s first find the angular frequency, ω, of the simple harmonic motion.
We know that the formula for the period, T, of a particle undergoing simple
harmonic motion is given by:
T=2π
ω
Given that T= 2 seconds, we can solve for ω:
2 = 2π
ω
ω=2π
2=πrad/s
Step 2: Next, let’s express the displacement, x, of the particle at any time,
t, using the general formula for simple harmonic motion:
x(t) = Asin(ωt +ϕ)
20
where: A= amplitude = 5 cm, ω= angular frequency = πrad/s, and ϕ=
phase constant.
Step 3: Since the particle is at its maximum displacement at time t= 0, we
have:
x(0) = Asin(ϕ)=5
sin(ϕ)=1
ϕ=π
2
Step 4: Now, we can find the displacement of the particle at time t= 1
second:
x(1) = 5 sinπ·1 + π
2
x(1) = 5 sinπ+π
2
x(1) = 5 sin3π
2
x(1) = 5 ·(−1) = −5 cm
Therefore, the displacement of the particle at time t= 1 second is −5 cm.
Question 26
Question
An object of mass mis attached to a spring with spring constant k. The object
is displaced from its equilibrium position by a small distance x0and released
from rest. Determine the period Tof the resulting simple harmonic motion in
terms of m,k, and x0.
Solution
Step 1: Apply Hooke’s Law to find the force exerted by the spring at a displace-
ment xfrom equilibrium.
Step 1: F=−kx
Step 2: Apply Newton’s Second Law to the object to set up the differential
equation for simple harmonic motion.
Step 2: md2x
dt2=−kx
Step 3: Solve the differential equation by assuming x(t) = Acos(ωt), where
Ais the amplitude of the motion and ωis the angular frequency.
Step 3: d2x
dt2=−ω2Acos(ωt)
21
m(−ω2Acos(ωt)) = −kA cos(ωt)
m(−ω2) = −k
ω=rk
m
Step 4: The period Tof the motion is related to the angular frequency ωby
T=2π
ω.
Step 4: T=2π
ω=2π
qk
m
= 2πrm
k
Therefore, the period Tof the simple harmonic motion is 2πpm
k.
Question 27
Question
A mass-spring system oscillates with an amplitude of 0.2 m and a period of 2
seconds. If the mass is 0.5 kg, determine the equation describing the position
of the mass as a function of time.
Solution
Step 1: Find the angular frequency of the oscillation. Given that the period
T= 2 seconds, we can use the formula T=2π
ωto find the angular frequency ω.
2 = 2π
ω
ω=2π
2=πrad/s
Step 2: Determine the equation for the position of the mass. The equation
for simple harmonic motion is given by:
x(t) = Asin(ωt +ϕ)
where: A= amplitude = 0.2 m, ω= angular frequency = πrad/s.
To find the phase constant ϕ, we need to consider the initial conditions.
When t= 0, x= 0.2 m (amplitude).
x(0) = 0.2=0.2 sin(ϕ)
sin(ϕ)=1
Since sin(90) = 1, we have ϕ= 90 = π
2.
Therefore, the equation describing the position of the mass as a function of
time is:
x(t)=0.2 sinπt +π
2
22
Question 28
Question
A particle of mass mis attached to a spring with spring constant k. If the
particle is displaced from its equilibrium position by a distance of Aand released
from rest, find the speed of the particle when it is at a distance A
2from the
equilibrium position.
Solution
Step 1: Find the angular frequency ωof the simple harmonic motion.
Given: m, k, A
The angular frequency ωof the simple harmonic motion is given by:
ω=rk
m
Step 2: Find the amplitude of the motion at the instant the speed is required.
Given: A, A
2
The amplitude of the motion at the instant the speed is required is A
2.
Step 3: Find the displacement xof the particle from the equilibrium point
at the instant the speed is required. The displacement xis given by:
x=Acos(ωt)
At the instant the speed is required, x=A
2.
A
2=Acos(ωt) =⇒cos(ωt) = 1
2
Step 4: Find the velocity of the particle at the instant the speed is required.
The velocity of the particle is given by:
v=−Aω sin(ωt)
Substitute the values of Aand ωinto the equation:
v=−Ark
msin rk
mt!
Step 5: Find the speed of the particle when it is at a distance A
2from the
equilibrium position. Substitute x=A
2into the equation for velocity:
v=−A
2rk
m
23
v=−1
2ωA =−1
2rk
mA
So, the speed of the particle when it is at a distance A
2from the equilibrium
position is 1
2qk
mA.
Question 29
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and a
period of 2 seconds. If the particle starts at a displacement of 3 cm to the right
of the equilibrium position at time t= 0, find the displacement equation for the
particle.
Solution
Step 1: Identify the general form of the displacement equation for simple har-
monic motion. The general form of the displacement equation for simple har-
monic motion with an amplitude Aand period Tis given by:
x(t) = Acos 2π
Tt+ϕ
where x(t) is the displacement of the particle at time t,ϕis the phase angle, A
is the amplitude, and Tis the period.
Step 2: Find the values of amplitude Aand period T. Given that the
amplitude A= 5 cm and the period T= 2 seconds.
Step 3: Find the phase angle ϕ. To find the phase angle ϕ, substitute the
initial conditions into the general form of the displacement equation:
x(0) = 5 cos ϕ= 3
Solving for ϕ:
cos ϕ=3
5
ϕ= cos−13
5
Step 4: Substitute the values of A,T, and ϕinto the displacement equation.
Therefore, the displacement equation for the particle is:
x(t) = 5 cos 2π
2t+ cos−13
5
x(t) = 5 cosπt + cos−13
5
24
Question 30
Question
A particle moves in simple harmonic motion with a period of 5 seconds. If at t
= 0, the displacement of the particle is 0.3 m and its velocity is 0.1 m/s in the
positive direction, find an expression for the displacement of the particle as a
function of time.
Solution
Given: Period of motion, T = 5 seconds
Amplitude, A = 0.3 m
Initial displacement, x(0) = 0.3 m
Initial velocity, v(0) = 0.1 m/s
We know that the general equation for simple harmonic motion is given by:
x(t) = Acos(ωt +ϕ)
where: - A is the amplitude, - is the angular frequency (equal to 2π
T), - is the
phase angle.
To find , we use the formula ω=2π
Twhere T is the period. Substituting T
= 5 seconds, we get:
ω=2π
5
To find the phase angle , we use the initial conditions given:
x(0) = Acos(ϕ)=0.3
v(0) = −Aω sin(ϕ)=0.1
Solving these equations simultaneously, we find .
Step 1: Find the angular frequency
ω=2π
5
Step 2: Find the phase angle From the given initial conditions:
Acos(ϕ)=0.3
−Aω sin(ϕ)=0.1
Dividing the second equation by the first:
−ωtan(ϕ) = 0.1
0.3
tan(ϕ) = −0.1
0.3ω
25
Solving for , we get:
ϕ= arctan −0.1
0.3ω
Therefore, the displacement function for the particle as a function of time
is:
x(t)=0.3 cos 2π
5t+ arctan −0.1
0.3·2π
5
x(t) = 0.3 cos 2π
5t+ arctan −5
9π
Question 31
Question
A particle of mass mis attached to a spring with spring constant k. It is
displaced from its equilibrium position by a distance 2Aand released from rest.
What is the maximum speed of the particle during the subsequent motion?
Solution
Step 1: Find the angular frequency ωof the oscillation.
The angular frequency ωis given by:
ω=rk
m
Step 2: Find the amplitude of the oscillation.
The amplitude Ais half of the total displacement, so in this case A=
1
2(2A) = A.
Step 3: Find the maximum speed of the particle.
The maximum speed of the particle occurs when the displacement is the
amplitude, i.e., x=A.
The velocity of the particle at any position xis given by:
v=ωpA2−x2
At x=A:
vmax =ωpA2−A2
vmax =ω·0
vmax = 0
Therefore, the maximum speed of the particle during the subsequent motion
is 0.
26
Question 32
Question
A particle of mass mis attached to a spring with spring constant k. At time
t= 0, the particle is displaced Aunits from its equilibrium position x= 0
and released from rest. Find an expression for the velocity of the particle as a
function of time.
Solution
Step 1: First, we can write the equation of motion for simple harmonic motion.
The force acting on the particle is given by Hooke’s Law as F=−kx, where x
is the displacement from the equilibrium position. Using Newton’s Second Law,
F=ma, we have
−kx =md2x
dt2
Step 2: Rearranging the equation, we get
d2x
dt2+k
mx= 0
Step 3: The general solution to this second-order homogeneous differential
equation is given by
x(t) = Acos(ωt) + Bsin(ωt)
where ω=qk
m.
Step 4: We know that at time t= 0, the particle has been displaced Aunits
from the equilibrium position x= 0 and released from rest, i.e., x(0) = Aand
dx
dt t=0 = 0. Substituting these initial conditions into the general solution, we
find
x(t) = Acos(ωt)
Step 5: To find the velocity of the particle as a function of time, we differ-
entiate the position function with respect to time:
dx
dt =−Aωsin(ωt)
Step 6: Therefore, the velocity of the particle as a function of time is given
by
v(t) = −Aωsin(ωt)
27
Question 33
Question
A particle undergoes simple harmonic motion with an amplitude of 4 cm and a
period of 3 seconds. If at t= 0, the particle is at its maximum displacement,
find the equation of motion for the particle.
Solution
Given that the particle is at its maximum displacement (amplitude) at t= 0,
we can assume that the initial displacement x0= 4 cm and the initial velocity
v0= 0.
The general equation of motion for simple harmonic motion is given by:
x(t) = Acos(ωt +ϕ)
where: - Ais the amplitude, - ωis the angular frequency, - ϕis the phase angle.
From the given information, we know that A= 4 cm and T= 3 seconds.
The angular frequency ωcan be found using the formula T=2π
ω:
ω=2π
T=2π
3radians/second
Therefore, the equation of motion for the particle is:
x(t) = 4 cos 2π
3t+ϕ
To find the phase angle ϕ, we can use the initial conditions x(0) = Aand
v(0) = 0.
x(0) = 4 cos(ϕ)=4
cos(ϕ) = 1
ϕ= 0 radians
So, the final equation of motion for the particle is:
x(t) = 4 cos 2π
3t
Question 34
Question
A particle undergoes simple harmonic motion with an amplitude of 5 cm and a
period of 4 seconds. If the particle is at its equilibrium position at t= 0 seconds
and moving in the positive direction, determine the displacement of the particle
at t= 1 second.
28
Solution
Step 1: Find the angular frequency ωusing the formula T=2π
ω, where Tis the
period. Step 2: Calculate the displacement at t= 1 second using the formula
x(t) = Acos(ωt), where x(t) is the displacement at time t. Step 3: Substitute
the values of A,ω, and tinto the formula to find the displacement.
Step 1: Given that the period T= 4 seconds, we use the formula
T=2π
ω
to find the angular frequency ω.
4 = 2π
ω=⇒ω=2π
4=π
2
Step 2: The displacement of the particle at any time tis given by
x(t) = Acos(ωt)
where Ais the amplitude. Given that the amplitude A= 5 cm, the equation
becomes
x(t) = 5 cos π
2t
Step 3: To find the displacement at t= 1 second, substitute t= 1 into the
equation.
x(1) = 5 cos π
2×1= 5 cos π
2= 5 ×0 = 0 cm
Therefore, the displacement of the particle at t= 1 second is 0 cm .
Question 35
Question
A block of mass mis attached to a spring with spring constant k. The block
is displaced from its equilibrium position and released. Determine the time for
the block to reach half of its amplitude for simple harmonic motion.
Solution
Let Abe the amplitude of the motion. The equation for the displacement xof
the block as a function of time tis given by x(t) = Asin(ωt), where ω=qk
m.
The time t1for the block to reach half of its amplitude can be obtained by
solving the equation x(t1) = A
2.
29
Step 1: Find the expression for t1
x(t1) = A
2
Asin(ωt1) = A
2
sin(ωt1) = 1
2
ωt1=π
6(for the first positive solution)
t1=π
6ω
Step 2: Substitute ωinto the expression for t1
t1=π
6ω
=π
6rm
k
Therefore, the time for the block to reach half of its amplitude is π
6pm
k.
30
