The amplitude of a particle executing S.H.M. is 4cm . At the mean position the speed of the particle is 16cms−1 . The distance of the particle from the mean position at which the speed of the particle becomes 8√3cms−1 , will be:

 Assume that the amplitude of S.H.M. be A , at mean position the velocity of the particle is maximum

hence, let velocity be Vmax ,

The velocity of the particle away from the mean position be V ,

Consider ω be the angular velocity of the particle and y be the position of the particle away from mean position.

Given data: Vmax =16cms−1 A=4cm V=8√3cms−1

Thus, let us use the formula of maximum velocity to find angular velocity such that

Vmax = Aω

⇒ 16 = 4×ω ⇒ω=4

Now, we have velocity of particle away from the mean position such that V=8√3cms−1

⇒V=ω√ A2−y2

⇒ 8√3 =4√42−y2

On squaring both the sides, we get ⇒(8√3)= 42(42−y2)

⇒ 8 × 8 × 3 = 4 × 4 × (16−y2)

⇒ 12 = 16−y2 On solving the above equation, we get

⇒ y = 2cm

Thus the particle position away from mean position is 2cm

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