This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Wave Motion I”.

1. A body of mass 5kg hangs from a spring and oscillates with a time period of 2π seconds. If the body is removed, the length of the spring will decrease by

a) g/k meters

b) k/g meters

c) 2π meters

d) g meters

View Answer

Explanation: Here

T=2π√(m/k)=2π or m/k=1

In equilibrium,

mg=kd

d=mg/k=1×g=g meters.

2. The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

a) T/4

b) T

c) T/2

d) 2T

View Answer

Explanation: T=2π√(m/k),k=F/x

When the spring is cut into four equal parts,

k

^{‘}=F/(x/4)=4(F/x)=4k

T

^{‘}=2π√(m/k

^{‘})=2π√(m/4k)=T/2.

3. Time period of a simple pendulum is 2sec. If its length is increased by 4 times, then its period becomes

a) 8 sec

b) 12 sec

c) 16 sec

d) 4 sec

View Answer

Explanation: T

^{‘}/T=√(l

^{‘}/l)=√(4l/l)=2

T

^{‘}=2T=2×2=4sec.

4. If the length of a simple pendulum is increased by 2%, then the time period

a) Increases by 1%

b) Decreases by 1%

c) Increases by 2%

d) Decreases by 2%

View Answer

Explanation: T∝√l

Percentage increase in time period,

∆T/T×100=1/2×∆l/l×100

=1/2×2%=1%.

5. A second’s pendulum is mounted in a rocket. Its period of oscillation will decrease when rocket is

a) Moving down with uniform acceleration

b) Moving around the earth in geostationary orbit

c) Moving up with uniform velocity

d) Moving up with uniform acceleration

View Answer

Explanation: When at rest, T=2π√(l/g)

When the rocket moves up with uniform acceleration,

T

^{‘}=2π√(l/(g+a))

Clearly, T

^{‘}<T.

6. A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by T=2π√(l/g), where g is equal to

a) g

b) g-a

c) g+a

d) √(g^{2}+a^{2} )

View Answer

Explanation: As g and a are acting along perpendicular directions, the effective value of acceleration due to gravity is

g

^{‘}=√(g

^{2}+a

^{2}).

7. In case of a forced vibration, the resonance peak becomes very sharp when the

a) Damping force is small

b) Restoring force is small

c) Applied periodic force is small

d) Quality factor is small

View Answer

Explanation: When the damping force is small, the resonance peak is high and narrow.

8. A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force Fsinωt. If the amplitude of the particle is maximum for ω=ω_{1} and the energy of the particle maximum for ω=ω_{2}, then

a) ω_{1}≠ω_{0} and ω_{2}=ω_{0}

b) ω_{1}=ω_{0} and ω_{2}=ω_{0}

c) ω_{1}≠ω_{0} and ω_{2}≠ω_{0}

d) ω_{1}≠ω_{0} and ω_{2}≠ω_{0}

View Answer

Explanation: In a driven harmonic oscillator, the energy is maximum at ω

_{2}=ω

_{0}and amplitude is maximum at frequency, ω

_{1}is lesser than ω

_{0}in the presence of a damping of force. Therefore, ω

_{1}≠ω

_{0}and ω

_{2}=ω

_{0}.

9. Two simple pendulums of lengths 5m and 20m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed

a) 2 oscillations

b) 1 oscillation

c) 5 oscillations

d) 3 oscillations

View Answer

Explanation: T

_{1}/T

_{2}=√(l

_{1}/l

_{2})=√(5/20)=1/2

T

_{1}=2T

_{2}

When the longer pendulum completes 2 oscillations, the shorter pendulum completes one oscillation and both are again in same phase.

10. The composition of two simple harmonic motions of equal periods at the right angle to each other and with a phase difference of π results in the displacement of the particle along

a) Circle

b) Figure of eight

c) Straight line

d) Ellipse

View Answer

Explanation: Let x = asinωt

y=bsin(ωt+π)=-bsinωt

x/a=y/b

y=-b/a×x

This is the equation of a straight line.

**Sanfoundry Global Education & Learning Series – Engineering Physics.**

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