A particle performing S.H.M. starts from the equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is π m s - 1 . Amplitude of oscillation is cos ⁡ 45 o = 1 √ 2

Question

A particle performing S.H.M. starts from the equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is π m s - 1 . Amplitude of oscillation is cos ⁡ 45 o = 1 √ 2, 2 2 m, 4 2 m, 6 2 m, 8 2 m

MARKS App · Accepted Answer

Displacement of the particle = x = A sin ⁡ ω t The velocity of the particle = v = d x d t = A ω cos ⁡ ω t v = π m s - 1 , T = 16 s , ω = 2 π T = 2 π 16 = π 8 rad s - 1 ∴ π = A × π 8 × cos ⁡ π 8 × 2 ∴ 1 = A 8 cos ⁡ π 4 = A 8 . 1 √ 2 ∴ A = 8 2 m

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