# find the point where th kienatic and potential energy is equal in SHM?

Arun
25750 Points
5 years ago

A simple realization of the harmonic oscillator in classical mechanics is a particle which is acted upon by a restoring force proportional to its displacement from its equilibrium position. Considering motion in one dimension, this means

F=−kx

Such a force might originate from a spring which obeys Hooke’s law

The force constant k is a measure of the stiffness of the spring. The variable x is chosen equal to zero at the equilibrium position, The negative sign shows restoring force, always in the opposite sense to the displacement x.

F = m .{d^2 x / dx^2} = −kx

where m is the mass of the body attached to the spring, which is itself assumed massless. This leads to a differential equation of familiar form, although with different variables:

the solution of the differential equation is of the form

x = A sin wt where

w= sqrt(k/m) and A is maximum displacement called amplitude,

if at t=0 x=0

thus the velocity at any time t will be

dx/dt = A.w.cos(wt)

so the kinetic energy at time t

(1/2) m. vel^2 = (1/2) m. {A^2. w^2 Cos^2 (wt)}

The general relation between force and potential energy in a conservative system in one dimension is

F = −(dV / dx) i.e. the force is derived as negative rate of change of potential

Thus the potential energy of a harmonic oscillator is given by

V(x) = (1/2). k .x ^2

which has the shape of a parabola,turning points ±x(max)

where the total energy E equals the potential energy (1/2). k .x ^2 while the kinetic energy is momentarily zero.

In contrast, when the oscillator moves past x=0 the kinetic energy reaches its maximum value while the potential energy equals zero.

therefore the two forms of energy are being shared with total E constant.

Now let at any instant t the two energies are equal

K.E. = P.E.

so (1/2) m. {A^2. w^2 Cos^2 (wt)} = (1/2). k .x(t) ^2 = (1/2). k .( A sin wt )^2

Both are equal to (1/2) total Energy

so its possible if wt= pi/4 where sin and cos function will have same value.

at wt= pi/4 the displacement x will be

x= a sin(wt) = A. sin (pi/4) = A. (1/sqrt(2) = A/1.414

x= 0.714 A

thus the kinetic energy and potential energy will be equal at x =0,714 A value as above