A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

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Aditi Chauhan · Answer

Sol. . r = 10cm Because, K.E. = P.E. So (1/2) m ω2 (r2– y2) = (1/2) m ω2y2 r2 – y2 = y2 ⇒ 2y2 = r2 ⇒ y = r/√2 = 10/√2 = 5√2 cm form the mean position

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A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

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A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

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