MHT CET · Physics · Oscillations
A mass \(0.4 \mathrm{~kg}\) performs S.H.M. with a frequency \(\frac{16}{\pi} \mathrm{Hz}\). At a certain displacement it has kinetic energy \(2 \mathrm{~J}\) and potential energy \(1.2 \mathrm{~J}\). The amplitude of oscillation is
- A 0.15 m
- B 0.125 m
- C 0.075 m
- D 0.1 m
Answer & Solution
Correct Answer
(B) 0.125 m
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{m}=0.4 \mathrm{~kg}, \mathrm{f}=\frac{16}{\pi} \mathrm{Hz}, \mathrm{K} \cdot \mathrm{E} \cdot=2 \mathrm{~J}, \text { P.E. }=1.2 \mathrm{~J} \\
& \omega=2 \pi \mathrm{f}=2 \pi \times \frac{16}{\pi}=32 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
Total energy T.E. \(=2+1.2=3.2 \mathrm{~J}\)
\(\mathrm{TE}=\frac{1}{2} \mathrm{~m}^2 \mathrm{~A}^2 \)
\( \therefore \mathrm{A}^2=\frac{2(\mathrm{TE})}{\mathrm{m} \omega^2}=\frac{2 \times 3.2}{0.4 \times(32)^2}=\) \(\frac{6.4}{0.4 \times(32)^2}=\frac{16}{(32)^2} \)
\( \therefore \mathrm{A}=\frac{4}{32}=\frac{1}{8}=0.125 \mathrm{~m}\)
\begin{aligned}
& \mathrm{m}=0.4 \mathrm{~kg}, \mathrm{f}=\frac{16}{\pi} \mathrm{Hz}, \mathrm{K} \cdot \mathrm{E} \cdot=2 \mathrm{~J}, \text { P.E. }=1.2 \mathrm{~J} \\
& \omega=2 \pi \mathrm{f}=2 \pi \times \frac{16}{\pi}=32 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
Total energy T.E. \(=2+1.2=3.2 \mathrm{~J}\)
\(\mathrm{TE}=\frac{1}{2} \mathrm{~m}^2 \mathrm{~A}^2 \)
\( \therefore \mathrm{A}^2=\frac{2(\mathrm{TE})}{\mathrm{m} \omega^2}=\frac{2 \times 3.2}{0.4 \times(32)^2}=\) \(\frac{6.4}{0.4 \times(32)^2}=\frac{16}{(32)^2} \)
\( \therefore \mathrm{A}=\frac{4}{32}=\frac{1}{8}=0.125 \mathrm{~m}\)
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