MHT CET · Physics · Oscillations
A simple pendulum of length \(L\) has mass \(m\) and it oscillates freely with amplitude A. At extreme position, its potential energy is ( \(\mathrm{g}=\) acceleration due to gravity)
- A \(\frac{m g A}{2 L}\)
- B \(\frac{\mathrm{mgA}^2}{\mathrm{~L}}\)
- C \(\frac{\mathrm{mgA}}{\mathrm{L}}\)
- D \(\frac{\mathrm{mgA}^2}{2 \mathrm{~L}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{mgA}^2}{2 \mathrm{~L}}\)
Step-by-step Solution
Detailed explanation
Potential energy of particle at extreme position is,
\(\begin{aligned}
\text {,P.E. } & =\frac{1}{2} m \omega^2 A^2 \\
& =\frac{1}{2} m \times \frac{g}{L} \times A^2 \quad \ldots\left(\because \omega=\sqrt{\frac{g}{l}}\right)
\end{aligned}\)
\(\begin{aligned}
\text {,P.E. } & =\frac{1}{2} m \omega^2 A^2 \\
& =\frac{1}{2} m \times \frac{g}{L} \times A^2 \quad \ldots\left(\because \omega=\sqrt{\frac{g}{l}}\right)
\end{aligned}\)
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