Q20 / 214 · NTA Abhyas (NEET) 2020 · Physics · Oscillations

A simple pendulum of length $L$ has mass $M$ and it oscillates freely with amplitude $A$ . At the extreme position, its potential energy is ( $g$ = acceleration due to gravity)

Correct answer
A
$\frac{M g A^{2}}{2 L}$
B
$\frac{M g A}{2 L}$
C
$\frac{M g A^{2}}{L}$
D
$\frac{2 M g A^{2}}{L}$

Solution

Potential energy

= \frac{1}{2} M ω^{2} A^{2}

= \frac{1}{2} M . \frac{g}{L} . A^{2} (∴ ω = \sqrt{\frac{g}{L}})

Hence, the correct answer is option: 'A'.

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