1.

The ratioof kinetic energy of mean position to thhe potential energy when the displacementis half of the amplitudeis :

Answer»

`(4)/(1)`
`(2)/(3)`
`(4)/(3)`
`(1)/(2)`.

SOLUTION :Kinetic energy `E_(K)=(1)/(2)m omega^(2)(r^(2)-y^(2))`
At MEAN position `y=0,:. E_(K)=(1)/(2)m omega^(2)r^(2)`
Potential energy `E_(P)=(1)/(2)m omega^(2)y^(2)`
At ` y=( r )/(2),E_(P)=(1)/(2)m omega^(2)(r^(2))/(4)`.
`:.""(E_(K))/(E_(P))=(4)/(1)`
Correct choice is (a).


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