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Question

A particle executes simple harmonic motion on a straight line. The amplitude of oscillation is 2 cm. At the instant when the displacement of the particle from mean position is 1 cm, the magnitude of acceleration is equal to the magnitude of velocity. The frequency of the simple harmonic motion is

A
37π Hz
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B
35π Hz
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C
32π Hz
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D
34π Hz
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Solution

The correct option is C 32π Hz
Given:
Amplitude of the oscillation, A=2 cm
Displacement of the particle from mean position, r=1 cm
It is given that when displacement of the particle from mean position is 1 cm, the magnitude of acceleration is equal to the magnitude of velocity of the particle.
a=v
ω2x=ωA2x2
ω(ωxA2x2)=0
ω=0 or ω=A2x2x
Angular frequency cannot be zero.
So, ω=A2x2x=22121=3 rad/s
Hence, the frequency of oscillation,
f=ω2π
f=32π Hz

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