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