JEE MAIN - Physics (2006 - No. 29)
Starting from the origin a body oscillates simple harmonically with a period of $$2$$ $$s.$$ After what time will its kinetic energy be $$75\% $$ of the total energy?
$${1 \over 6}s$$
$${1 \over 4}s$$
$${1 \over 3}s$$
$${1 \over 12}s$$
Paliwanag
$$K.E.\,$$ of a body undergoing $$SHM$$ is given by,
$$K.E. = {1 \over 2}m{a^2}{\omega ^2}{\cos ^2}\,\omega t,$$
$$T.E. = {1 \over 2}m{a^2}{\omega ^2}$$
Given $$K.E.=0.75T.E.$$
$$ \Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = {\pi \over 6}$$
$$ \Rightarrow t = {\pi \over {6 \times \omega }} \Rightarrow t = {{\pi \times 2} \over {6 \times 2\pi }} \Rightarrow t = {1 \over 6}s$$
$$K.E. = {1 \over 2}m{a^2}{\omega ^2}{\cos ^2}\,\omega t,$$
$$T.E. = {1 \over 2}m{a^2}{\omega ^2}$$
Given $$K.E.=0.75T.E.$$
$$ \Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = {\pi \over 6}$$
$$ \Rightarrow t = {\pi \over {6 \times \omega }} \Rightarrow t = {{\pi \times 2} \over {6 \times 2\pi }} \Rightarrow t = {1 \over 6}s$$