MHT CET · Physics · Oscillations
A particle is suspended from a vertical spring which is executing S.H.M. of frequency \(5 \mathrm{~Hz}\). The spring is unstretched at the highest point of oscillation. Maximum speed of the particle is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)\)
- A \(\frac{1}{\pi} \mathrm{m} / \mathrm{s}\)
- B \(\frac{1}{4 \pi} \mathrm{m} / \mathrm{s}\)
- C \(\frac{1}{2 \pi} \mathrm{m} / \mathrm{s}\)
- D \(\pi \mathrm{m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\pi} \mathrm{m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Amplitude of SHM, \(\mathrm{A}=\frac{\mathrm{mg}}{\mathrm{k}}=\frac{\mathrm{g}}{\omega^2}\)
\(
\begin{aligned}
& \omega=2 \pi \mathrm{f}=2 \pi \times 5=10 \pi \\
& \therefore \mathrm{A}=\frac{10}{100 \pi^2}=\frac{1}{10 \pi^2} \\
& \mathrm{~V}_{\max }=\mathrm{A} \omega=\frac{1}{10 \pi^2} \times 10 \pi=\frac{1}{\pi} \mathrm{m} / \mathrm{s}
\end{aligned}
\)
\(
\begin{aligned}
& \omega=2 \pi \mathrm{f}=2 \pi \times 5=10 \pi \\
& \therefore \mathrm{A}=\frac{10}{100 \pi^2}=\frac{1}{10 \pi^2} \\
& \mathrm{~V}_{\max }=\mathrm{A} \omega=\frac{1}{10 \pi^2} \times 10 \pi=\frac{1}{\pi} \mathrm{m} / \mathrm{s}
\end{aligned}
\)
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