Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2432 Upvotes
Verified Answer
The correct answer is:
$6\ V$
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
$\begin{aligned}
\mathrm{V}^{\prime} & =\mathrm{A}^{\prime} \omega^{\prime} \\
& =(2 \mathrm{~A}) \times(3 \omega) \\
& =6 \mathrm{~A} \omega \\
& =6 \mathrm{~V}
\end{aligned}$
$\therefore \quad \omega^{\prime}=\frac{2 \pi}{\mathrm{T}^{\prime}}=\frac{2 \pi}{\left(\frac{1}{3} \mathrm{~T}\right)}=\frac{6 \pi}{\mathrm{T}}=3 \omega$
$\therefore \quad$ The new maximum velocity
$\begin{aligned}
\mathrm{V}^{\prime} & =\mathrm{A}^{\prime} \omega^{\prime} \\
& =(2 \mathrm{~A}) \times(3 \omega) \\
& =6 \mathrm{~A} \omega \\
& =6 \mathrm{~V}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.