If a particle under S.H.M. has a time period | vedclass

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(A) The maximum velocity $(v_{\max})$ of a particle performing $S.H.M.$ is given by the formula $v_{\max} = a\omega$.Here,$a$ is the amplitude and $\o

Vedclass · Accepted Answer

$\frac{\pi}{25} \ m/s$

Vedclass · Answer

(A) The maximum velocity $(v_{\max})$ of a particle performing $S.H.M.$ is given by the formula $v_{\max} = a\omega$.Here,$a$ is the amplitude and $\omega$ is the angular frequency.The angular frequency $\omega$ is related to the time period $T$ by $\omega = \frac{2\pi}{T}$.Given: Amplitude $a = 2 	imes 10^{-3} \ m$ and Time period $T = 0.1 \ s$.Substituting these values into the formula:$v_{\max} = a 	imes \frac{2\pi}{T} = (2 	imes 10^{-3}) 	imes \frac{2\pi}{0.1}$.$v_{\max} = \frac{4 	imes 10^{-3} 	imes \pi}{10^{-1}} = 4 	imes 10^{-2} 	imes \pi = 0.04\pi \ m/s$.$v_{\max} = \frac{4\pi}{100} = \frac{\pi}{25} \ m/s$.Therefore,the correct option is $A$.

If a particle under $S.H.M.$ has a time period of $0.1 \ s$ and an amplitude of $2 \times 10^{-3} \ m$,what is its maximum velocity?

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