A particle performing S.H.M. with maximum velocity ' $V$ '. If the amplitude double and periodic time is…

A particle performing S.H.M. with maximum velocity ' $V$ '. If the amplitude double and periodic time is made, $\left(\frac{1}{3}\right)^{\text {rd }}$ then the maximum velocity is
  1. $\frac{\mathrm{V}}{2}$
  2. $\frac{\mathrm{V}}{3}$
  3. 6V
  4. $\frac{2 \mathrm{~V}}{3}$

Solution

Given: - The amplitude doubles $\left(A^{\prime}=2 A\right)$. - The period becomes one-third $\left(T^{\prime}=\frac{1}{3} T\right)$.
Formula for Maximum Velocity: $V_{\max }=A \omega$ where $\omega=\frac{2 \pi}{T}$. Step 1: New Angular Velocity: $\omega^{\prime}=\frac{2 \pi}{T^{\prime}}=\frac{2 \pi}{\frac{1}{3} T}=6 \pi / T=3 \omega$
Step 2: New Maximum Velocity: $V_{\max }^{\prime}=A^{\prime} \omega^{\prime}=(2 A)(3 \omega)=6 A \omega$
Since the original $V_{\max }=A \omega$ : $V_{\max }^{\prime}=6 V$

Asked in: MHT CET 2024 (10 May Shift 1)

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