Question 335: Oscillations

Question

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

Accepted Answer

At extreme position, the potential energy of simple pendulum,
$P E=\frac{1}{2} m A^2 \omega^2 \quad 	ext{.... (i)}$
where, $\omega=$ angular frequency $=\frac{2 \pi}{T}$
For simple pendulum, $T=2 \pi \sqrt{\frac{L}{g}}$
$	herefore \quad \omega=\frac{2 \pi}{2 \pi} \sqrt{\frac{g}{L}}=\sqrt{\frac{g}{L}} \quad\left[\because T=\frac{2 \pi}{\omega}ight]$
On putting value of $\omega$ in Eq. (i), we get
$\mathrm{PE}=\frac{1}{2} m A^2\left(\sqrt{\frac{g}{L}}ight)^2=\frac{m g A^2}{2 L}$

Answer

$\frac{m g A}{L}$

Answer

$\frac{m g A}{2 l}$

Answer

$\frac{m g A^2}{L}$

Answer

$\frac{m g A^2}{2 L}$

Oscillations

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

Choose the correct answer: