The displacement $x$ (in $metres$) of a particle performing simple harmonic motion is related to time $t$ (in $seconds$) as $x = 0.05\cos{\left( 4\pi t + \frac{\pi}{4} \right).\ }$ The frequency of the motion will be

Question

Dr. Abubacker · Accepted Answer

(d) From the given equation $\omega = 2\pi n = 4\pi \Rightarrow n = 2Hz$

Q. The displacement $x$ (in $metres$) of a particle performing simple harmonic motion is related to time $t$ (in $seconds$) as $x = 0.05\cos{\left( 4\pi t + \frac{\pi}{4} \right).\ }$ The frequency of the motion will be

Explanation

Reviewed by Dr. Abubacker

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