Question

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)

पर्याय

• ${\displaystyle \frac{2{\text{mgA}}^2}{\text{L}}}$

• ${\displaystyle \frac{{\text{mgA}}^2}{\text{L}}}$

• ${\displaystyle \frac{{\text{mgA}}^2}{2\text{L}}}$

• ${\displaystyle \frac{\text{mgA}}{2\text{L}}}$

Answer 1

A simple pendulum of length 'L' has mass 'm' and it oscillates freely with amplitude 'A'. At the extreme position, its potential energy is ${\displaystyle \underline{\frac{mg{A}^2}{2L}}}$.

 (g = acceleration due to gravity)

Explanation:

PE = ${\displaystyle \frac{1}{2}\text{m}{\omega }^2{\text{A}}^2}$ = ${\displaystyle \frac{1}{2}{\text{kA}}^2}$ = ${\displaystyle \frac{1}{2}\frac{\text{mg}}{\text{L}{\text{A}}^2}}$ = ${\displaystyle \frac{{\text{mgA}}^2}{2\text{L}}}$

F = -kx mgsinθ = -kA k = ${\displaystyle \frac{\text{mg}\theta }{\text{A}}}$ = ${\displaystyle \frac{\text{mgA}}{\text{LA}}}$