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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=ω√A2−x2 ⇒ω(ωx−√A2−x2)=0 ⇒ω=0 or ω=√A2−x2x Angular frequency cannot be zero. So, ω=√A2−x2x=√22−121=√3 rad/s Hence, the frequency of oscillation, f=ω2π ⇒f=√32π Hz

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