Question

Obtain expressions for kinetic energy, potential energy and total energy of a particle performing linear S.H.M.

Answer 1

Acceleration of the particle , performing S.H.M is given by $\alpha =-{\omega }^{2}y$

where $\omega$ is the angular velocity, and y is the displacement of particle.

now, workdone by particle = $\overrightarrow{F}.\overrightarrow{d}y$

as we know, acceleration and displacement are in opposite directions in case of S.H.M

so, $W=-m{\omega }^{2}ydy$

where m is the mass of the particle.

$W=-m{\omega }^2\int ydy$

$W=-\frac{1}{2}m{\omega }^2{y}^2$

so, potential energy = -W

$=\frac{1}{2}m{\omega }^2{y}^2$

we know, $\omega =2\pi \eta$

so, $P.E=2{\pi }^2{\eta }^{2}m{y}^2$ ......(1)

velocity of particle , $v=\omega Acos\omega t$

or, $v=\omega \sqrt{A^2-y^2}$

so, kinetic energy of particle, $K.E=\frac{1}{2}m{v}^2$

hence, $K.E=\frac{1}{2}m{\omega }^2(A^2-y^2)$

but $\omega =2\pi \eta$

so, $K.E=2{\pi }^2{\eta }^{2}m(A^2-y^2)$ ....(2)

so, total mechanical energy = K.E + P.E

$=2{\pi }^2{\eta }^{2}m{A}^2$ ......(3)