Question

(c
Show that the total energy of the particle
performing linear S.H.M. is constant.​

Answer 1

Heya !

We know,

Total energy of a particle performing linear S.H.M is equal to K.E. and P.E.,

where K.E. is attained due to velocity of particle and P.E. is due to the displacement of particle from the mean position.

K.E.

K.E. = $\frac{1}{2}$ m$v^{2}$

we know,

v = aωcosωt

Now,

K.E. = $\frac{1}{2}$ m $a^{2}$ ω^2 $cos^{2}$ ωt

      = $\frac{1}{2}$ m $a^{2}$ ω^2 ( 1 - $sin^{2}$ωt )

      = $\frac{1}{2}$ m $a^{2}$ ω^2 ( 1 - $\frac{y^{2} }{a^{2} }$ )                                           [y=asinωt]

      = $\frac{1}{2}$ m $a^{2}$ ω^2 ( $\frac{a^{2} - y^{2} }{a^{2} }$ )

      = $\frac{1}{2}$ m ω^2 ($a^{2} - y^{2}$)

P.E.

We know,

Restoring force  (F) ∝ y (displacement)

i.e. ,

F = - ky   [ (-)ve because restoring force acts in direction opposite to mean position ]  

Let  the particle be displaced through dy

So,

Work done ⇒    

dW = - Fdy

dW = kydy

W = $\int\limits^y_0 {} \, dW$ = $\int\limits^y_0 {ky} \, dy$

W = $\int\limits^y_0 {} \, ky^{2}$ × $\frac{1}{2}$

W = $\frac{1}{2}$ k $y^{2}$

   = $\frac{1}{2}$ m ω^2 $y^{2}$

Now,

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

          =   $\frac{1}{2}$ m ω^2 ($a^{2} - y^{2}$) +  $\frac{1}{2}$ m ω^2 $y^{2}$

         = $\frac{1}{2}$ m ω^2 $a^{2}$

So, clearly we can see that the T.E. is independent of the position of particle during S.H.M. , so it remains constant .

Thank you !