The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

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Solution

The correct option is A
Maximum acceleration $= a w^{2} = a \times 4 π^{2} n^{2}$
$= 0.01 \times 4 \times (π)^{2} \times (60)^{2} = 144 π^{2} m / s e c^{2}$

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The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

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