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Karnataka Board PUCPUC Science Class 11

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

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Question

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

Options

  • 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)`.

MCQ
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Solution

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)`.

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Chapter 14: Oscillations - Exercises [Page 98]

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NCERT Exemplar Physics [English] Class 11
Chapter 14 Oscillations
Exercises | Q 14.6 | Page 98

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