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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 l0 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 =A32= 23 cm
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