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The displacement x (in metre) of a particle in, simple harmonic motion is related to time t (in seconds) as
$x = 0 .01 \cos (π t + \frac{π}{4})$
The frequency of the motion will be

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0.5 Hz

1.0 Hz

π2Hz

π Hz

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detailed solution

Correct option is A

Comparing given equation with standard equation,x=acos (ωt+φ) we get, a=0.01 and ω=π ⇒2πn=π⇒n=0.5 Hz

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The displacement x (in metre) of a particle in, simple harmonic motion is related to time t (in seconds) as x=0.01cosπ t+π4The frequency of the motion will be

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The displacement x (in metre) of a particle in, simple harmonic motion is related to time t (in seconds) as
$x = 0 .01 \cos (π t + \frac{π}{4})$
The frequency of the motion will be