#### A particle of mass executes simple harmonic motion with amplitude   and frequency  . The average kinetic energy during its motion from the  position of equilibrium to the end is Option 1) Option 2) Option 3) Option 4)

As we learnt in

Potential energy in S.H.M. -

- wherein

Where

For a particle to execute simple harmonic motion its displacement at any time  t is given by  $x(t)=a(cos\, \omega t+\phi )$

where, $a= amplitude,\omega =angular \; frequency,\phi =phase \; constant.$

$Let \; us \; choose\: \phi =0$

$\therefore \; \; x(t)=a\: cos\: \omega t$

Velocity of a particle $\nu = \frac{dx}{dt}= -a\: \omega \: \sin \omega t$

$Kinetic \: energy\: of \: a \: particle\: is\: K= \frac{1}{2} m\nu ^{2}$

$= \frac{1}{2}ma^{2}\omega ^{2}\sin ^{2}\omega t$

$Average\: kinetic \: energy\: < K>$

$= < \frac{1}{2}ma^{2}\omega ^{2}\sin ^{2}\omega t>$

$= \frac{1}{2}m\omega ^{2}a^{2}< \sin ^{2}\omega t>$

$= \frac{1}{2}m\omega ^{2}a^{2}\left ( \frac{1}{2} \right )$

$\left [ \because \sin ^{2}\Theta > = \frac{1}{2} \right ]$

$= \frac{1}{4}ma^{2}\left ( 2\pi \upsilon \right )^{2}\: \: \: \: \: \: \: \left [\because \omega = 2\pi \upsilon \right ]$

$= \pi ^{2}ma^{2}\upsilon$

Correct option is 2.

Option 1)

This is an incorrect option.

Option 2)

This is the correct option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.