JEE MAIN - Physics (2007 - No. 26)
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
$$2{\pi ^2}\,m{a^2}{v^2}$$
$${\pi ^2}\,m{a^2}{v^2}$$
$${1 \over 4}\,m{a^2}{v^2}$$
$$4{\pi ^2}m{a^2}{v^2}$$
Explanation
KEY CONCEPT : The instantaneous kinetic energy of a particle executing $$S.H.M.$$ is given by
$$K = {1 \over 2}m{a^2}{\omega ^2}{\sin ^2}\omega t$$
$$\therefore$$ average $$K.E. = < K > = < {1 \over 2}m{\omega ^2}{a^2}{\sin ^2}\omega t > $$
$$ = {1 \over 2}m\omega {}^2{a^2} < {\sin ^2}\omega t > $$
$$ = {1 \over 2}m{\omega ^2}{a^2}\left( {{1 \over 2}} \right)$$
$$\left( \, \right.$$ as $$\left. { < {{\sin }^2}\theta > = {1 \over 2}} \right)$$
$$ = {1 \over 4}m{\omega ^2}{a^2} = {1 \over 4}m{a^2}{\left( {2\pi v} \right)^2}$$
$$\left( \, \right.$$ $$\left. {\omega = 2\pi v} \right)$$
or, $$\,\,\,\,\, < K > = {\pi ^2}m{a^2}{v^2}$$
$$K = {1 \over 2}m{a^2}{\omega ^2}{\sin ^2}\omega t$$
$$\therefore$$ average $$K.E. = < K > = < {1 \over 2}m{\omega ^2}{a^2}{\sin ^2}\omega t > $$
$$ = {1 \over 2}m\omega {}^2{a^2} < {\sin ^2}\omega t > $$
$$ = {1 \over 2}m{\omega ^2}{a^2}\left( {{1 \over 2}} \right)$$
$$\left( \, \right.$$ as $$\left. { < {{\sin }^2}\theta > = {1 \over 2}} \right)$$
$$ = {1 \over 4}m{\omega ^2}{a^2} = {1 \over 4}m{a^2}{\left( {2\pi v} \right)^2}$$
$$\left( \, \right.$$ $$\left. {\omega = 2\pi v} \right)$$
or, $$\,\,\,\,\, < K > = {\pi ^2}m{a^2}{v^2}$$
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