A particle is performing S.H.M. with maximum | vedclass

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Question

(A) The maximum velocity of a particle performing $S.H.M.$ is given by the formula: $V_{max} = \omega A = \frac{2 \pi}{T} A$.Here,$A$ is the amplitude

Vedclass · Accepted Answer

$1.5 v$

Vedclass · Answer

(A) The maximum velocity of a particle performing $S.H.M.$ is given by the formula: $V_{max} = \omega A = \frac{2 \pi}{T} A$.Here,$A$ is the amplitude and $T$ is the periodic time.Given that the initial maximum velocity is $v = \frac{2 \pi A}{T}$.If the amplitude is tripled $(A' = 3A)$ and the periodic time is doubled $(T' = 2T)$,the new maximum velocity $V'_{max}$ is:$V'_{max} = \frac{2 \pi A'}{T'} = \frac{2 \pi (3A)}{2T} = \frac{3}{2} \left( \frac{2 \pi A}{T} \right) = 1.5 v$.Therefore,the new maximum velocity will be $1.5 v$.

$A$ particle is performing $S.H.M.$ with maximum velocity $v$. If the amplitude is tripled and periodic time is doubled,then the new maximum velocity will be:

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