The amplitude of a particle executing simple harmonic motion with a frequency of `60` Hz is `0.01` m. Determine the maximum value of the acceleration of the particle.
Text Solution
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Maximum accleration `a_("max") = omega^(2)A = 4pi^(2)n^(2)A = 4pi^(2)n^(2)A = 4pi^(2)(60)^(2) xx (0.01) = 144 pi^(2)m//s^(2)`
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