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 $\alpha=-\omega^2y$
where $\omega$ is the angular velocity, and y is the displacement of particle.

now, workdone by particle = $\vec{F}.\vec{dy}$
as we know, acceleration and displacement are in opposite directions in case of S.H.M
so, $W=-m\omega^2y.dy$
where m is the mass of the particle.
$W=-m\omega^2\int{y}dy$
$W=-\frac{1}{2}m\omega^2y^2$

so, potential energy = -W
= $\frac{1}{2}m\omega^2y^2$
we know, $\omega=2\pi\eta$

so, $P.E =2\pi^2\eta^2my^2$......(1)

velocity of particle , $v=\omega Acos\omega t$
or, $v=\omega\sqrt{A^2-y^2}$
so, kinetic energy of particle, K.E = 1/2 mv²

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

so, total mechanical energy = K.E + P.E
= $2\pi^2\eta^2mA^2$......(3)

Answer 2

pata hi nahi hai to kay answer du yrr