Maximum speed of a particle in simple harmoni | vedclass

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(D) The maximum speed of a particle in simple harmonic motion is given by $v_{max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The maximum speed of a particle in simple harmonic motion is given by $v_{max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the amplitude.Since $\omega = \frac{2\pi}{T}$,we have $v_{max} = \frac{2\pi A}{T}$,which implies $A = \frac{v_{max} T}{2\pi}$.In one complete oscillation,the particle travels a total distance of $4A$ in time $T$.The average speed is defined as the total distance divided by the total time: $	ext{Average Speed} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{4A}{T}$.Substituting the expression for $A$ into the average speed formula: $	ext{Average Speed} = \frac{4}{T} 	imes \left( \frac{v_{max} T}{2\pi} ight) = \frac{4 v_{max}}{2\pi} = \frac{2v_{max}}{\pi}$.

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

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