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Q.

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

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a

52π

b

4π5

c

2π3

d

5π

answer is B.

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Detailed Solution

Amplitude A = 3 cmWhen particle is at x = 2 cm,its |velocity| = |acceleration|i.e., ωA2-x2 = ω2x ⇒ ω = A2-x2xT = 2πω = 2π(25) = 4π5

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