# Engineering Physics Questions and Answers – Harmonic Motion – 1

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Harmonic Motion – 1”.

1. A motion can be oscillatory but not simple harmonic.
a) True
b) False

Explanation: A motion can be oscillatory but not simple harmonic =. When a ball is dropped from a height on a perfectly elastic surface, the motion is oscillatory but not simple harmonic as restoring force is constant.

2. A particle executes simple harmonic motion between x = -A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then?
a) T1 < T2
b) T1 > T2
c) T1=T2
d) T1=2T2

Explanation: x=Asinωt
For x=A/2,
sinωT1=1/2
ωT1=π/6 or T1=π/2ω
For x=A,
sinω(T1+T2)=1
T1+T2=π/2ω
T2=π/2ω-π/6ω=π/3ω=2T1.

3. A particle free to move along the x-axis has potential energy given by U(x) = k [1-exp -x2] for -∞≤x≤+∞, where k is a positive constant of appropriate dimensions. Then?
a) The particle is in unstable equilibrium at points away from the origin
b) There is a force directed away from the origin for any finite nonzero value of x
c) It has its minimum kinetic energy at the origin if its total mechanical energy is k/2
d) The motion is simple harmonic for small displacement from x=0

Explanation: U=k(1-e(-x2))
F=-dU/dx=-2kxe(-x2)=-2kx(1-x2+⋯)
For small x, F ≅-2kx
This shows that the force is directed towards the origin and for smaller x, F∝x. Hence the motion is simple harmonic.

4. A simple pendulum has a time period T1, when on the earth’s surface; and T2, when taken to a height R above the earth’s surface (R is the radius of the earth). The value of T2/T1 is?
a) 1
b) √2
c) 4
d) 1

Explanation: g/g=(R/(R+R))2=1/4
As
T∝1/√g
T2/T1 = √(g/g)=√(4/1)=2.

5. A coin is placed on a horizontal platform, which undergoes vertical simple harmonic motion of angular frequency ω. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time.
a) For an amplitude of g22
b) For an amplitude of g/ω2
c) At the highest position of the platform
d) At the mean position of the platform

Explanation: The coin will remain in contact with the platform if amax does not exceed g, that is amax is at the most equal to g.
amax=g
2=g
a=g/ω2.

6. A child swinging on a swing stands up. Then the time period of the swing will ___________
a) Increase
b) Decrease
c) Remain the same
d) Increase, if the child is long and decreases if the child is short

Explanation: The child and the swing together constitute a pendulum of a time period,
T∝2π√(l/g)
As the girl stands up, her centre of gravity is raised. The distance between the point of suspension and the centre of gravity decreases, that is the length l decreases. Hence the time period T decreases.

7. If a particle is oscillating on the same horizontal plane in the ground ___________
a) It has only kinetic energy but no potential energy
b) It has only potential energy but no kinetic energy
c) It has both kinetic and potential energies
d) It has neither kinetic nor potential energies

Explanation: The particle has both kinetic and potential energies.

8. If the length of simple pendulum is tripled, what will be its new time period in terms of original period T?
a) 0.7T
b) 1.73T
c) T/2
d) T

Explanation: T∝√l
T∝√3l
T=√3l=1.732T.

9. Statement: The time period of a pendulum on a satellite orbiting the earth is infinity.
Reason: The time period of a pendulum is inversely proportional to square to root of g.
a) Both statement and reason are true and the reason is the correct explanation of the statement
b) Both statement and reason are true but the reason is not the correct explanation of the statement
c) Statement is true, but the reason is false
d) Statement and reason are false

Explanation: Both the statement and reason are true but reason is not the correct explanation of the statement. In satellite, g=0 and hence the time period is infinity.

10. A spring 40mm long is stretched by the acceleration of a force. If 10N force is required to stretch the spring through 1mm, then work done in stretching through 40mm is?
a) 84J
b) 48J
c) 24J
d) 8J

Explanation: k=F/x=10N/1mm=10N/(10-3m)=(104)N/m
Work done in stretching the spring through 40mm,
W=1/2×kx2=1/2×104×0.04×0.04=8J.

Sanfoundry Global Education & Learning Series – Engineering Physics.

To practice all areas of Engineering Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]