Question

A particle is executing simple harmonic motion with amplitude A. When the ratio of its kinetic energy to the potential energy is `1/4`, its displacement from its mean position is ______.

Options

• `2/sqrt5`A

• `sqrt3/2`A

• `3/4`A

• `1/4`A

Answer 1

A particle is executing simple harmonic motion with amplitude A. When the ratio of its kinetic energy to the potential energy is `1/4`, its displacement from its mean position is `underlinebb(2/sqrt5"A")`.

Explanation:

As we know

kinetic energy = `1/2"m"omega^2("A"^2-"x"^2)`

potential energy = `1/2"m"omega^2"x"^2`

∴ `(1/2"m"omega^2("A"^2-"x"^2))/(1/2"m"omega^2"x"^2)` = `1/4`

⇒ `("A"^2-"x"^2)/"x"^2` = `1/4`

4A² - 4x² = x²

⇒ x² = `4/5`A²

⇒ x =`2/sqrt5`A.