Question 358: 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)

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