Maximum speed of a particle in simple harmoni | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{2v_{max}}{\pi}$

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(D) The displacement of a particle in simple harmonic motion is $x(t) = A \sin(\omega t)$.The velocity is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t)$.The maximum speed is $v_{max} = A\omega$.The average speed is defined as the total distance traveled divided by the total time taken.In one complete oscillation,the particle travels from $x = 0$ to $x = A$,then to $x = -A$,and back to $x = 0$.The total distance traveled is $d = A + 2A + A = 4A$.The time taken for one complete oscillation is the time period $T = \frac{2\pi}{\omega}$.Therefore,the average speed $v_{avg} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{4A}{T} = \frac{4A}{2\pi/\omega} = \frac{2A\omega}{\pi}$.Substituting $v_{max} = A\omega$,we get $v_{avg} = \frac{2v_{max}}{\pi}$.

Maximum speed of a particle in simple harmonic motion is $v_{max}.$ Then average speed of a particle in one complete oscillation is equal to

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