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

• `(2"mgA"^2)/"L"`

• `("mgA"^2)/"L"`

• `("mgA"^2)/(2"L")`

• `("mgA")/(2"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 `underlinebb((mgA^2)/(2L))`.

 (g = acceleration due to gravity)

Explanation:

PE = `1/2"m"omega^2"A"^2` = `1/2"kA"^2` = `1/2 ("mg")/("L" "A"^2)` = `("mgA"^2)/(2"L")`

F = -kx mgsinθ = -kA k = `("mg" theta)/"A"` = `("mgA")/("LA")`