A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Question

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is, 5 π, 5 2 π, 4 π 5, 2 π 3

MARKS App · Accepted Answer

At position x from the mean position in S.H.M. velocity, v = ω A 2 - x 2 acceleration, a = x ω 2 here ω is the angular frequency and A is amplitude. Given in the problem, v = a ⇒ ω A 2 - x 2 = x ω 2 ⇒ 3 2 - 2 2 = 2 2 π T 5 = 4 π T Time period, T = 4 π 5

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