Question

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance `(2 A)/3` from equilibrium position. The new amplitude of the motion is ______.

Options

• `A sqrt 3`

• `(7 A)/3`

• `A/3 sqrt 41`

• 3A

Answer 1

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance `(2 A)/3` from equilibrium position. The new amplitude of the motion is `bbunderline((7 A)/3)`.

Explanation:

In S.H.M., velocity is given by:

v = `omega sqrt (A^2 - x^2)`

At x = `(2 A)/3`:

v₁ = `omega sqrt (A^2 - ((2 A)/3)^2)`

= `omega sqrt (A^2 - (4 A^2)/9)`

= `omega sqrt ((5 A^2)/9)`

= `(omega A sqrt 5)/3`

If the speed is tripled:

v₂ = 3 × v₁

= `3 xx ((omega A sqrt 5)/3)`

= `omega A sqrt 5`

The angular frequency (ω) remains the same. Use the velocity formula again with the new amplitude:

v₂ = `omega sqrt ((A')^2 - x^2)`

`omega A sqrt 5 = omega sqrt ((A')^2 - ((2 A)/3)^2)`    ...[Squaring on both sides and cancelling ω²]

5 A² = `(A')^2 - (4 A^2)/9`

(A')² = `5 A^2 + (4 A^2)/9`

= `(45 A^2 + 4 A^2)/9`

(A')² = `(49 A^2)/9`

A' = `sqrt ((49 A^2)/9)`

= `(7 A)/3`