Question

The displacement of a particle varies with time according to the relation y = a sin ωt + b cos ωt.

विकल्प

• The motion is oscillatory but not S.H.M.

• The motion is S.H.M. with amplitude a + b.

• The motion is S.H.M. with amplitude a² + b².

• The motion is S.H.M. with amplitude `sqrt(a^2 + b^2)`.

Answer 1

The motion is S.H.M. with amplitude `sqrt(a^2 + b^2)`.

Explanation:

According to the question, the displacement

y = a sin ωt + b cos ωt

Let a = A sin `phi` and b = A cos `phi`

Now, a² + b² = A² sin² `phi` + A² cos² `phi`

= A²

⇒ A = `sqrt(a^2 + b^2)`

y = A sin `phi` sin ωt + A cos `phi` cos ωt

= A sin (ωt + `phi`)

`(dy)/(dt) = Aω cos (ωt + phi)`

`(d^2y)/(dt^2) = - Aω^2 sin(ωt + phi)`

= `- Ayω^2`

= `(- Aω^2)y`

⇒ `(d^2y)/(dt^2) ∝ (-y)`

Hence, it is an equation of SHM with amplitude A = `sqrt(a^2 + b^2)`.