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Q.

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is

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a

3 cm

b

5 cm

c

2(3) cm

d

2(5) cm

answer is C.

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Detailed Solution

vmax=aω ⇒ω=vmaxa=104Now, v=ωa2−y2 ⇒v2=ω2(a2−y2) ⇒y2=a2−v2ω2⇒y=a2-v2ω2=42-5210/42=23cm

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