Simple Harmonic Motion Test

Result

Simple Harmonic Motion Test - 12

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Question 1
1 / -0

A ring of radius R is hung by a nail on its periphery such that it can freely rotate in its vertical plane. The time period of the ring for small oscillations is:

A

B

C

D

Solution
Question 2
1 / -0

In a stationary lift a spring-block system oscillates with a frequency f. When the lift accelerates up the frequency becomes f', then

A
f' > f

B
f' < f

C
f' = f

D
Any of the above depending on the value of the acceleration of the lift

Solution

The time period of the spring-block system does not depend on the effective acceleration due to gravity. So, the frequency also does not depend and remains the same.
Question 3
1 / -0

The equation of a SHM is given as x = 5 sin(4πt + π/3), where t is in seconds and x in meters. During a complete cycle, the average speed of the oscillator is:

A
Zero

B
10 m/s

C
20 m/s

D
40 m/s

Solution
Question 4
1 / -0

If a simple pendulum is brought deep inside a mine from earth surface, its time period of oscillation will

A
Increase

B
Decrease

C
Remain the same

D
Any of the above depending on the length of the pendulum

Solution

As the effective acceleration due to gravity decreases, the time period of oscillations of the simple pendulum increases.
Question 5
1 / -0

The initial phase of the particle executing SHM with y = 4 sin ωt + 3 cos ωt is

A
53°

B
37°

C
90°

D
45°

Solution
Question 6
1 / -0

Two simple harmonic motions are represented by y₁ = 6 sin ( 2πt + π/3) and y₂ = 3(sin 2πt + cos 2πt). The ratio of their amplitudes is

A
√2

B
2/3

C
1/2

D
2√2

Solution
Question 7
1 / -0

The graph between velocity and acceleration of a particle executing S.H.M. can be

A
A circle

B
An ellipse

C
A straight line

D
Both (1) & (2)

Solution
Question 8
1 / -0

The equation of simple harmonic motion is given by X = (4 cm)sin (6.25 t + 5π/3), then maximum velocity of the particle in simple harmonic motion is:

A
25.12 m/s

B
25.12 cm/s

C
12.56 m/s

D
12.56 cm/s

Solution
Question 9
1 / -0

The time period of a simple pendulum in a stationary trolley is T₁. If the trolley is moving with a constant speed, then time period is T₂, then

A
T₁ > T₂

B
T₁ < T₂

C
T₁ = T₂

D
T₂ = ∞T

Solution

The time period will remain the same as the acceleration of the trolley is zero in both the cases and the effective acceleration of pendulum remains the same.
Question 10
1 / -0

A particle executes S.H.M with a frequency of 20 Hz. The frequency with which its potential energy oscillates is:

A
5 Hz

B
20 Hz

C
10 Hz

D
40 Hz

Solution

The frequency at which potential energy oscillates= 2 x frequency of oscillation

= 2 x 20

= 40 Hz.