Simple Harmonic Motion Test

Result

Simple Harmonic Motion Test - 13

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Question 1
1 / -0

A body of mass 20 g is executing S.H.M with amplitude 5 cm. When it passes through equilibrium position its speed is 20 cm/s. Find the distance from equilibrium when its speed becomes 10 cm/s.

A

B

C

D

Solution

Hint: Find relation between velocity and displacement

Step 1: Find angular frequency, ω

Step 3:

Find x when velocity is 10 cm/s
Question 2
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For a simple harmonic oscillator, a velocity-time diagram is shown. The angular frequency of oscillation is:

A
24/2 rad/s

B
25π/4 rad/s

C
25 rad/s

D
25π rad/s

Solution

From the diagram:
Question 3
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A particle executing S.H.M. Its displacement x varies with time t as x = 8sin4t + 6cos4t. The maximum velocity of the particle will be:

A
20 unit

B
10 unit

C
40 unit

D
5 unit

Solution
Question 4
1 / -0

Which of the following statements is true for an ideal simple pendulum?

A
The expression for the time period is true only for the oscillation of infinitely small amplitude.

B
The motion of the bob is rotatory.

C
The buoyancy of air has an appreciable effect on the motion of the bob.

D
All of these

Solution

Only statement (1) is true as the expression is for only small oscillations and the bob does not rotate about its axis. Also, the air does not have a significant effect.
Question 5
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If a bob of the simple pendulum having negative charge q is made to oscillate in a uniform electric field acting vertically upward direction, then its time period:

A
Increases

B
Decreases

C
No effect of the electric field

D
Decreases or increases

Solution

When there is an electric field in an upward direction, the pendulum will experience a force in the downward direction.
Question 6
1 / -0

Find the percentage change in time period if the length of a simple pendulum is increased by 3%

A
3%

B
6%

C
1.5%

D
2.5%

Solution
Question 7
1 / -0

A particle is executing SHM with time period T. The time taken by it to travel from mean position to 1/√2 times its amplitude is equal to

A
T/6

B
T/12

C
T/8

D
T/4

Solution
Question 8
1 / -0

The motion of the particle is started at t = 0 and the equation of motion is given by x = 8 sin(100t + π/6), where x is in cm and t is in seconds. When will the particle come to rest for the first time?

A

B

C

D

Solution
Question 9
1 / -0

The time period of a body under S.H.M. is T₁ and T₂when restoring forces F₁ and F₂ respectively act on it. What will be the time period of S.H.M. when both the forces act simultaneously on it?

A

B
T₁ + T₂

C

D

Solution
Question 10
1 / -0

A particle is subjected to two mutually perpendicular SHM such that x = 2sinωt and y = 2 sin [ωt + π/2]. The path of the particle will be

A
An ellipse

B
A straight line

C
A parabola

D
A circle

Solution