Question

Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path

Answer 1

we know, standard equation of simple harmonic motion, $y=Asin\omega t$
so, velocity of particle of shm, $v=\omega Acos\omega t$ or, $v=\omega\sqrt{A^2-y^2}$.....(1)
and acceleration, a = -$\omega^2Asin\omega t$ or, $a=-\omega^2y$

potential energy of simple harmonic oscillator :
Workdone = F.dy = |ma| |dy| cos180°
=- m$\omega^2y$dy
= -1/2 m$\omega^2$y²
so, potential energy = negative of workdone
= 1/2 m$\omega^2$y²

kinetic energy = 1/2 mv²
= 1/2 m$\omega^2(A^2-y^2)$ [ from equation (1),


now, total energy = kinetic energy + potential energy
= 1/2 m$\omega^2y^2$ + 1/2m\omega^2(A^2-y^2)[/tex]
= 1/2 m$\omega^2A^2$

so, total energy is independent from displacement of particle , y.

hence, it is clear that total energy of a particle in simple harmonic motion is constant at any point on its path

Answer 2

Answer:

Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path