Question
A particle of mass $$m$$ executes simple harmonic motion with amplitude $$a$$ and frequency $$\nu .$$The average kinetic energy during its motion from the position of equilibrium to the end is
A.
$$2{\pi ^2}m{a^2}{\nu ^2}$$
B.
$${\pi ^2}m{a^2}{\nu ^2}$$
C.
$$\frac{1}{4}m{a^2}{\nu ^2}$$
D.
$$4{\pi ^2}m{a^2}{\nu ^2}$$
Answer :
$${\pi ^2}m{a^2}{\nu ^2}$$
Solution :
KEY CONCEPT: The instantaneous kinetic energy of a particle executing S.H.M. is given by
$$\eqalign{
& K = \frac{1}{2}m{a^2}{\omega ^2}{\sin ^2}\omega t \cr
& \therefore {\text{average}}\,K.E. = < K > = < \frac{1}{2}m{\omega ^2}{a^2}{\sin ^2}\omega t > \cr
& = \frac{1}{2}m{\omega ^2}{a^2} < {\sin ^2}\omega t > \cr
& = \frac{1}{2}m{\omega ^2}{a^2}\left( {\frac{1}{2}} \right)\,\left( {\because < {{\sin }^2}\theta > = \frac{1}{2}} \right) \cr
& = \frac{1}{4}m{\omega ^2}{a^2} = \frac{1}{4}m{a^2}{\left( {2\pi \nu } \right)^2}\left( {\because \omega = 2\pi \nu } \right) \cr
& {\text{or,}}\, < K > = {\pi ^2}m{a^2}{\nu ^2} \cr} $$