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(C) For a particle executing $SHM$,the maximum velocity is given by $V_{\max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the ampli

Vedclass · Accepted Answer

$6.28$

Vedclass · Answer

(C) For a particle executing $SHM$,the maximum velocity is given by $V_{\max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the amplitude.The maximum acceleration is given by $a_{\max} = \omega^2 A$.According to the problem,the magnitudes are equal: $V_{\max} = a_{\max}$.Substituting the expressions: $\omega A = \omega^2 A$.Dividing both sides by $\omega A$ (assuming $A 
eq 0$ and $\omega 
eq 0$),we get $\omega = 1 	ext{ rad/s}$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.Substituting $\omega = 1$,we get $T = 2\pi \approx 2 	imes 3.14 = 6.28 	ext{ s}$.

If the maximum velocity and maximum acceleration of a particle executing $SHM$ are equal in magnitude,the time period will be .... $\sec$

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