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A particle executes simple harmonic moti...

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?

Text Solution

Verified by Experts

The correct Answer is:
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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