Question
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
A.
$$2{\pi ^2}m{a^2}{v^2}$$
B.
$${\pi ^2}m{a^2}{v^2}$$
C.
$$\frac{1}{4}m{a^2}{v^2}$$
D.
$$4{\pi ^2}m{a^2}{v^2}$$
Answer :
$${\pi ^2}m{a^2}{v^2}$$
Solution :
The kinetic energy of a particle executing $$S.H.M.$$ is given by
$$\eqalign{
& K = \frac{1}{2}m{a^2}{\omega ^2}{\sin ^2}\omega t \cr
& = < \frac{1}{2}m{\omega ^2}{a^2}{\sin ^2}\omega t > = \frac{1}{2}m{\omega ^2}{a^2} < {\sin ^2}\omega t > \cr
& = \frac{1}{2}m{\omega ^2}{a^2}\left( {\frac{1}{2}} \right) \cr
& = \frac{1}{4}m{\omega ^2}{a^2}\,\,\left( {\because < {{\sin }^2}\theta > = \frac{1}{2}} \right) \cr
& = \frac{1}{4}m{a^2}\left( {2\pi {v^2}} \right)\,\,\left( {\because \omega = 2\pi v} \right) \cr
& {\text{or,}}\,\, < K > = {\pi ^2}m{a^2}{v^2} \cr} $$