Physics - SHM Question with Solution | TestHub

The maximum velocity of a particle in S.H.M. is given by $V=A\omega =\frac{2\pi A}{T}$, where $A$ is the amplitude and $T$ is the time period.

Given that the new amplitude is $A^{\prime }=2A$ and the new time period is $T^{\prime }=\frac{T}{3}$.

The new maximum velocity $V^{\prime }$ is:

$V^{\prime }=\frac{2\pi A^{\prime }}{T^{\prime }}$

Substituting the values of $A^{\prime }$ and $T^{\prime }$:

$V^{\prime }=\frac{2\pi (2A)}{\frac{T}{3}}$

$V^{\prime }=6\left(\frac{2\pi A}{T}\right)$

$V^{\prime }=6V$

Answer: 6V