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 :
