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Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. A particle of mass m executes simple harmonic motion with amplitude 'a' and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is :

Answer»

`(1)/(4)ma^(2)v^(2)`
`4pi^(2)ma^(2)v^(2)`
`2pi^(2)ma^(2)v^(2)`
`PI ma^(2)v^(2)`.

Solution :The average VALUE of kinetic energy during motion from mean to EXTREME POSITION `=(1)/(2)((1)/(2)m omega^(2)a^(2)+0)`
`=(1)/(4)m(2pi v)^(2).a^(2)=pi ma^(2)v^(2)`
So correct CHOICE is (d).


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