# Engineering Physics Questions and Answers – Wave Motion – 2

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This set of Engineering Physics online test focuses on “Wave Motion – 2”.

1. A particle in simple harmonic motion is described by the displacement function x(t)=Acos⁡(ωt+θ). If the initial (t=0) position of the particle is 1cm and its initial velocity isπcm/s, what is its amplitude? The angular frequency is the particle is πrad/s.
a) 1 cm
b) √2 cm
c) 2 cm
d) 2.5 cm

Explanation: v=ω√(A2-x2)
π=π√(A2-1)
A2-1=1 or A2=2
A=√2cm.

2. A particle executes simple harmonic motion, its time period is 16s. If it passes through the centre of oscillation, then its velocity is 2 m/s at time 2s. The amplitude will be ___________
a) 7.2m
b) 4cm
c) 6cm
d) 0.72m

Explanation: Here,
t=2s,v=2m/s,T=16s
v=Aωcosωt
2=A×2π/14×cos⁡(2π/16×2)
A=(16√2)/π=7.2m.

3. A body is executing the simple harmonic motion with an angular frequency of 2rad/sec. Velocity of the body at 20m displacement, when amplitude of motion is 60m, is ___________
a) 90 m/s
b) 118 m/s
c) 113 m/s
d) 131 m/s

Explanation: v=ω√(A2-y2)=2√(602-202)
v=80√2
v=113m/s.

4. A particle is executing simple harmonic motion of amplitude 10cm. Its time period of oscillation is π seconds. The velocity of the particle when it is 2 cm from extreme position is ___________
a) 10 cm/s
b) 12 cm/s
c) 16√16 cm/s
d) 16 cm/s

Explanation: v=2π/T×√(A2-y2)
v=2π/π √(102-82)
=2×6=12cm/s.

5. The magnitude of acceleration of particle executing simple harmonic motion at the position of maximum displacement is?
a) Zero
b) Minimum
c) Maximum
d) Infinity

Explanation: Acceleration in simple harmonic motion is
a=ω2 y
At ymax=A, amax2 A
Acceleration is maximum at the position of maximum displacement.

6. The maximum velocity and maximum acceleration of a body moving in a simple harmonic motion are 2m/s and 4m/s2 respectively. What will be the angular velocity?

Explanation: vmax=ωA, amax2 A
ω=amax/vmax = 4/2

7. A particle executing simple harmonic motion has amplitude 0.01 and frequency 60Hz. The maximum acceleration of the particle is ___________
a) 144 π2 m/s2
b) 80 π2 m/s2
c) 120 π2 m/s2
d) 60 π2 m/s2

Explanation: amax2 A=4π2 v2 A
=4π2×60×60×0.01=144 π2 m/s2.

8. A particle having potential energy 1/3 of the maximum value at a distance of 4 cm from mean position. Amplitude of motion is ___________
a) 4√3
b) 6/√2
c) 2/√6
d) 2√6

Explanation: Ep=1/3 E
1/2 ky2=1/3×1/2×kA2
A=√3 y=√3×4=4√3 cm.

9. A particle executes simple harmonic motion of amplitude A. At what distance from the mean position is its kinetic energy equal to its potential energy?
a) 0.51A
b) 0.71A
c) 0.61A
d) 0.81A

Explanation: Ek=Ep
1/2 k(A2-y2)=1/2×ky2
y=±A/√2
y=±0.71A.

10. To show that a simple pendulum executes simple harmonic motion, it is necessary to assure that ___________
a) Length of the pendulum is small
b) Amplitude of oscillation is small
c) Mass of the pendulum is small
d) Acceleration due to gravity is small

Explanation: Motion is simple harmonic only when the amplitude of oscillation is small because only then f is proportional to x.

11. Time period of a simple pendulum will be double if we ___________
a) Decrease the length 2 times
b) Decrease the length 4 times
c) Increase the length 2 times
d) Increase the length 4 times

Explanation: T=2π√(l/g) or T∝√l
When the length is increased four times, time period gets doubled.

12. The time period of a simple pendulum is 2 sec. If its length is increased by 4 times, then its period becomes ___________
a) 16 sec
b) 8 sec
c) 12 sec
d) 4 sec

Explanation: T∝√l, T∝√4l
T/T=2
T=2T=2×2=4 sec.

13. A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21%, the increase in the time period of the pendulum of increased length is?
a) 10%
b) 30%
c) 21%
d) 50%

Explanation: As T is proportional to l, the percentage increase in a time period on increasing the length by 21%
=1/2×∆l/l×100=1/2×21=10.5%.

14. A hollow spherical pendulum is filled with mercury has time period T. If mercury is thrown out completely, then the new time period ___________
a) Increases
b) Decreases
c) Remains the same
d) First increases and then decreases

Explanation: Position of centre of gravity remains unaffected when mercury is thrown out. Hence effective length and time period remain the same.

15. A simple pendulum is vibrating in an evacuated chamber. It will oscillate with ___________
a) Constant amplitude
b) Increasing amplitude
c) Decreasing amplitude
d) First increasing amplitude and then decreasing amplitude

Explanation: In a vacuum, there is no loss of energy due to resistive forces. So, amplitude remains constant.

16. Two simple pendulum, whose lengths are 100cm and 121cm, are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again?
a) 11
b) 10
c) 21
d) 20

Explanation: T∝√l
For the two pendulums in same phase
nT1=(n+1) T2
n√(l1)=(n+1) √(l2)
n√121=(n+1)√100
n×11=(n+1)10
n=10.

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