# Energy in SHM , Kinetic Energy , Potential Energy

#### Kinetic Energy (K.E)

From definition of kinetic energy,
$\displaystyle K = \frac{1}{2}mv^2$

Let , x = Asinωt

On differentiating with respect to time ,

$\displaystyle \frac{dx}{dt} = \omega A cos\omega t$

Hence , v = ω A cosωt

$\displaystyle K = \frac{1}{2}m\omega^2 A^2 cos^2\omega t$

$\displaystyle K = \frac{1}{4}m\omega^2 A^2 (1 + cos2\omega t)$

Kinetic energy varies periodically with double the frequency of SHM

$\displaystyle v = \omega (\sqrt{A^2 – x^2} )$

$\displaystyle K = \frac{1}{2}m\omega^2 (A^2 – x^2 )$

for x = 0,

$\displaystyle K = \frac{1}{2}m\omega^2 A^2$ = Kmax

for x = A, K = 0 = Kmin

#### Potential Energy (PE)

Potential Energy U is given by

$\displaystyle \int dU = -\int dW = -\int \vec{F}.\vec{dx}$

$\displaystyle U = -\int_{0}^{x}m\omega^2 x dx cos180$

$\displaystyle U = \frac{1}{2}m\omega^2 x^2$

In above derivation

(i) Angle between F and dx is taken as 180° as the two are oppositely directed.

(ii) Reference zero for U is taken at x = 0.

$\displaystyle U = \frac{1}{2}m\omega^2 A^2 sin^2\omega t$

$\displaystyle U = \frac{1}{4}m\omega^2 A^2 (1 – cos2\omega t)$

NOTE : Like Kinetic energy, Potential energy also varies periodically with double the frequency of SHM

$\displaystyle U = \frac{1}{2}m\omega^2 x^2$

at x = 0, U = 0 = Umin

for x = A ,

$\displaystyle U = \frac{1}{2}m\omega^2 A^2$

### Total Mechanical Energy in SHM

Total Mechanical Energy , E = K + U = $\displaystyle U = \frac{1}{2}m\omega^2 A^2$ = constant

The variation of K , U and E as a function of displacement and time are shown below.

If y = A sin ωt ;

then $\displaystyle K = \frac{1}{4}m\omega^2 A^2 (1 + cos2\omega t)$

and $\displaystyle U = \frac{1}{4}m\omega^2 A^2 (1 – cos2\omega t)$

=> E = K + U = $\displaystyle \frac{1}{2}m\omega^2 A^2$ = constant

i.e. E does not depend upon time or x.

The variation of K, U and E as function of position and time are shown below :