The maximum velocity of a particle performing S.H.M. is ' $V$ '. If the periodic time is made…
The maximum velocity of a particle performing S.H.M. is ' $V$ '. If the periodic time is made $\left(\frac{1}{3}\right)^{\mathrm{rt}}$ and the amplitude is doubled, then the new maximum velocity of the particle will be
${\frac {V}{6}}$
${\frac {3V}{2}}$
$3\ V$
$6\ V$
Solution
Given $\mathrm{T}^{\prime}=\frac{1}{3} \mathrm{~T}$ and $\mathrm{A}^{\prime}=2 \mathrm{~A}$
$\therefore \quad \omega^{\prime}=\frac{2 \pi}{\mathrm{T}^{\prime}}=\frac{2 \pi}{\left(\frac{1}{3} \mathrm{~T}\right)}=\frac{6 \pi}{\mathrm{T}}=3 \omega$
$\therefore \quad$ The new maximum velocity
$\begin{aligned}
\mathrm{V}^{\prime} & =\mathrm{A}^{\prime} \omega^{\prime} \\
& =(2 \mathrm{~A}) \times(3 \omega) \\
& =6 \mathrm{~A} \omega \\
& =6 \mathrm{~V}
\end{aligned}$
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