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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 = 10cm =0.1m kinetic energy = mA2w2 cos2wt/2 ................1 potential energy = mA2w2sin2wt/2 ..................2 PE = KE (GIVEN) so tanwt = 1 sinwt =1/sqrt(2) now , x = Asinwt =A/sqrt2= 7cm
A = 10cm =0.1m
kinetic energy = mA2w2 cos2wt/2 ................1
potential energy = mA2w2sin2wt/2 ..................2
PE = KE (GIVEN)
so
tanwt = 1
sinwt =1/sqrt(2)
now , x = Asinwt
=A/sqrt2= 7cm
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