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प्रश्न
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)`.
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उत्तर
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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