MHT CET · Physics · Oscillations
A particle performing linear S.H.M. of amplitude \(0.1 \mathrm{~m}\) has displacement \(0.02 \mathrm{~m}\) and acceleration \(0.5 \mathrm{~m} / \mathrm{s}^2\). The maximum velocity of the particle in \(\mathrm{m} / \mathrm{s}\) is
- A 0.05
- B 0.5
- C 0.01
- D 0.25
Answer & Solution
Correct Answer
(B) 0.5
Step-by-step Solution
Detailed explanation
Acceleration \(\mathrm{a}=\omega^2 \mathrm{x}\)
\(
\begin{aligned}
& \therefore \omega^2=\frac{\mathrm{a}}{\mathrm{x}}=\frac{0.5}{0.02}=25 \\
& \therefore \omega=5 \mathrm{rad} / \mathrm{s} \\
& \mathrm{V}_{\max }=\mathrm{A} \omega=0.1 \times 5=0.5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \omega^2=\frac{\mathrm{a}}{\mathrm{x}}=\frac{0.5}{0.02}=25 \\
& \therefore \omega=5 \mathrm{rad} / \mathrm{s} \\
& \mathrm{V}_{\max }=\mathrm{A} \omega=0.1 \times 5=0.5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
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