MHT CET · Physics · Oscillations
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
- A \(\frac{\mathrm{V}}{2}\)
- B \(\frac{\mathrm{V}}{3}\)
- C 6V
- D \(\frac{2 \mathrm{~V}}{3}\)
Answer & Solution
Correct Answer
(C) 6V
Step-by-step Solution
Detailed explanation
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 \(V_{\max }=A \omega\) :
\(V_{\max }^{\prime}=6 V\)
- 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 \(V_{\max }=A \omega\) :
\(V_{\max }^{\prime}=6 V\)
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