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.

1.	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?
Answer» Correct Answer - B::C `r=10cm ` Because K.E. =P.E. `=so, (1/2)momega^2(r^2-y^2)=(1/2)momega^2y^2` `r^2-y^2=y^2` `2y^2=r^2` `rarr y=r/sqrt2=10/sqrt2 =5sqrt2` from 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?

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