Obtain expressions for the instantaneous kinetic energy potential energy & total energy of a simple harmonic oscillator.

Question collected from Telegram Group

Solution:

As the restoring force acts in the simple harmonic motion, the particle executing simple harmonic motion has both Potential Energy and Kinetic Energy.

When there is no damping on oscillation, then total energy is equal to the sum of Potential Energy and Kinetic Energy.

i.e. T.E. = P.E. + K.E…(i)

For S.H.M. the acceleration can be expressed as

     a = -$\omega $2y

Then restoring force is given by

F = ma = -m$\omega $2y…….. (2)

Let ‘dy’ be the small displacement then small work done ‘dW’ is given by

dW= -F.dy….. (3)     [-ve sign indicates that the work is done against the restoring force.]

Then total work done can be obtained by integrating equation (3). i.e. 

$\int\limits_{0}^{y}{dw}$=    $\int\limits_{0}^{y}{F\,dy}$

Or, W =$\int\limits_{0}^{y}{(m{{\omega }^{2}}y)dy}$ 

Or, W = m${{\omega }^{2}}$ $\int\limits_{0}^{y}{y\,dy}$

Or, W = m$\omega $2 $\left[ \frac{{{y}^{2}}}{2} \right]$$\underset{0}{\overset{y}{\mathop{{}}}}\,$

Or, W = m$\omega $2  [$\frac{{{y}^{2}}}{2}$ – $\frac{{{0}^{2}}}{2}$]

Or, W = $\frac{1}{2}$ m$\omega $2 y2

According to the work energy theorem this work done is stored as P.E.

$\therefore $       P.E. = $\frac{\mathbf{1}}{\mathbf{2}}$ m$\omega $2y2……… (4)

This is the required expression for P.E. of simple harmonic oscillator.

Now, K.E. = $\frac{1}{2}$ mv2

Or, K. E. = $\frac{1}{2}$ m [$\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$]2   $\because $ v = $\omega $$\sqrt{{{r}^{2}}-{{y}^{2}}}$ 

$\therefore $      K.E. = $\frac{\mathbf{1}}{\mathbf{2}}$ m$\omega $2 (r2 – y2) ……. (5)

This is the required expression for K.E. of simple harmonic oscillator.

From (1),

T.E. = P.E. + K.E.

Or, T.E. = $\frac{1}{2}$ m$\omega $2y2 + $\frac{1}{2}$ m$\omega $2 (r2 – y2)

 Or, T.E. = $\frac{1}{2}$ m$\omega $2 y2 + $\frac{1}{2}$ m$\omega $2r2 – $\frac{1}{2}$ m$\omega $2y2

$\therefore $          T.E. = $\frac{\mathbf{1}}{\mathbf{2}}$ m$\omega $2 r2        

This is the required expression for total energy of a simple harmonic oscillator.

Special Cases:

(i) At mean position, y = 0

P.E.= $\frac{1}{2}$ m$\omega $2y2 = 0

K.E = $\frac{1}{2}$ m$\omega $2 (r2 – y2) = $\frac{1}{2}$ m$\omega $2r2 = T.E.

(ii) At extreme positions, y = r

K.E = $\frac{1}{2}$ m$\omega $2 (r2 – y2) = 0

P.E.= $\frac{1}{2}$ m$\omega $2y2 = $\frac{1}{2}$ m$\omega $2r2 = T.E  

The K.E. and P.E. change with position. When one increases the other decreases but the total energy of the body in S.H.M. remains constant. The variation of energy is shown in figure below