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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vikas askiitian expert · Answer

A = 10cm =0.1m kinetic energy = mA 2 w 2 cos 2 wt/2 ................1 potential energy = mA 2 w 2 sin 2 wt/2 ..................2 PE = KE (GIVEN) so tanwt = 1 sinwt =1/sqrt(2) now , x = Asinwt =A/sqrt2= 7cm

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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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