Simple Harmonic Motion Test

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Simple Harmonic Motion Test - 19

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Question 1
1 / -0

The displacement of a particle executing S.H.M. is given by x = 0.01sin100π(t + 0.05). The time period is:

A
0.01 s

B
0.02 s

C
0.1 s

D
0.2 s

Solution
Question 2
1 / -0

A boy is swinging in a swing. If he stands, the time period will:

A
First decrease, then increase

B
Decrease

C
Increase

D
Remain same

Solution

The time period is given by:

L → the length of the string (or) the distance from top to the centre of mass of the man.

When the man stands up, the length decreases and the time period also decreases.
Question 3
1 / -0

The time period of a simple pendulum in a freely falling lift will be:

A
Finite

B
Infinite

C
Zero

D
All of these

Solution

The time period is given by:

g_eff → effective acceleration due to gravity.

g_eff is zero in a freely falling lift. So the time period tends to infinity.
Question 4
1 / -0

If the effective length of a simple pendulum is equal to the radius of the earth (R), the time period will be:

A

B

C

D

Solution

For a pendulum with an effective length l comparable to the radius of the Earth R, the time period T is given by:
Question 5
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A body executing S.H.M. along a straight line has a velocity of 3 ms^-1 when it is at a distance of 4 m from its mean position and 4 ms^-1 when it is at a distance of 3 m from its mean position. Its angular frequency and amplitude are:

A
2 rad s^-1 & 5 m

B
1 rad s^-1& 10 m

C
2 rad s^-1& 10 m

D
1 rad s^-1& 5 m

Solution

Velocity,
Question 6
1 / -0

The frequency of oscillation of a mass m suspended by a spring is ν₁. If the length of the spring is cut to one third, then the same mass oscillates with a frequency ν_2,then:

A
ν₂= 3ν₁

B
3ν₂ = ν₁

C
ν₂ = √3ν₁

D
√3ν₂ = v₁

Solution
Question 7
1 / -0

The plot of velocity (v) versus displacement (x) of a particle executing simple harmonic motion is shown in the figure. The time period of osciliation of the particle is:

A
π/2 s

B
π s

C
2π s

D
3π s

Solution
Question 8
1 / -0

The equation of simple harmonic motion may not be expressed as (each term has its usual meaning):

A
x = A sin (ωt + ϕ)

B
x = A cos (ωt + ϕ)

C
x = a sin ωt + b cos ωt

D
x = A sin (ωt+ϕ) + B sin (2ωt+ϕ)

Solution

Simple harmonic motion can be represented by a single sine or cosine function.

cannot be presented as a single sine function.
Question 9
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If a particle is executing simple harmonic motion, then the acceleration of the particle:

A
is uniform.

B
varies linearly with time.

C
is non-uniform.

D
Both (2) & (3)

Solution

The acceleration of a particle performing simple harmonic motion is given by A = −ω²x

The acceleration varies along with the displacement lineary.
Question 10
1 / -0

What is the phase difference between the acceleration and the velocity of a particle executing simple harmonic motion?

A
Zero

B
π/2

C
π

D
2π

Solution