If the frequency of a simple harmonic motion | vedclass

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(C) The displacement of a particle in simple harmonic motion is given by $x = A \sin(\omega t)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\om

Vedclass · Accepted Answer

$2f$

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(C) The displacement of a particle in simple harmonic motion is given by $x = A \sin(\omega t)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m(A \omega \cos(\omega t))^2 = \frac{1}{2}mA^2 \omega^2 \cos^2(\omega t)$.Using the trigonometric identity $\cos^2(	heta) = \frac{1 + \cos(2	heta)}{2}$,we get:$K = \frac{1}{2}mA^2 \omega^2 \left( \frac{1 + \cos(2\omega t)}{2} ight) = \frac{1}{4}mA^2 \omega^2 (1 + \cos(2\omega t))$.The frequency of the kinetic energy is determined by the term $\cos(2\omega t)$.Since the angular frequency of the kinetic energy is $2\omega$,the frequency $f'$ is given by $f' = \frac{2\omega}{2\pi} = 2 \left( \frac{\omega}{2\pi} ight) = 2f$.

If the frequency of a simple harmonic motion is $f$,what is the frequency of its kinetic energy?

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