1.

The angular velocity and the amplitude of a simple pendulum are omega and a, respectively. The ratio of its kinetic and potential energies at a displacement x from the mean position is……………

Answer»

`(X^(2)omega^(2))/(a^(2)-x^(2)omega^(2))`
`(x^(2))/(a^(2)-x^(2))`
`(a^(2)-x^(2)omega^(2))/(x^(2)omega^(2))`
`(a^(2)-x^(2))/(x^(2))`

SOLUTION :Kinetic ENERGY `K=(1)/(2)momega^(2)(a^(2)-x^(2))`
Potential energy `U=(1)/(2)momega^(2)x^(2)implies(K)/(U)=(a^(2)-x^(2))/(x^(2))`


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