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 ends is

Options

A. `2p^(2)mA^(2)upsilon^(2)`
B. `p^(2)mA^(2)upsilon^(2)`
C. `(1)/(4)pi^(2)mA^(2)upsilon^(2)`
D. `4pi^(2)mA^(2)upsilon^(2)`

Answer

A

Explanation

Kinetic energy of the particle executing SHM at an instant when its displacemnt from the mean position is x is `K=(1)/(2)momega^(2)(A^(2)-x^(2))`
At the mean position, `x=0,K_(1)=(1)/(2)momega^(2)A^(2)`
At the extreme position, `x=A,K_(2)=(1)/(2)momega^(2)(A^(2)-A^(2))=0`
`therefore` Average kinetic energy = `(K_(1)+K_(2))/(2)=(((1)/(2)momega^(2)A^(2)+0))/(2)`
`=(1)/(4)momega^(2)A^(2)=(1)/(4)m(4pi^(2)upsilon^(2))A^(2)(because omega=2piupsilon)=pi^(2)mA^(2)upsilon^(2)`