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Show that for a Particle in Linear Shm the Average Kinetic Energy Over a Period of Oscillation Equals the Average Potential Energy Over the Same Period. - Physics

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Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

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Solution 1

The equation of displacement of a particle executing SHM at an instant is given as:

`x =Asin omegat`


A = Amplitude of oscillation

ω = Angular frequency = `sqrt(k/M)`

The velocity of the particle is:

`v = (dx)/(dt) = Aomegacosomegat`

The kinetic energy of the particle is:

`E_k = 1/2 Mv^2 = 1/2 MA^2omega^2 cos^2 omegat`

 The potential energy of the particle is:

`E_rho = 1/2 kx^2= 1/2 Momega^2 A^2 sin^2 omegat`

For time period T, the average kinetic energy over a single cycle is given as:

`(E_k)_"Avg" = 1/T int_0^T E_k dt`

= `1/T int_0^T 1/2 MA^2 omega^2 cos^2 omega t dt`

= `1/2T MA^2 omega^2 int_0^T (1+cos 2 omegat)/2 dt`

`= 1/(4T) MA^2omega^2[t + (sin 2 omegat)/(2omega)]_0^T`

`= 1/(4T) MA^2 omega^2 (T)`

`= 1/4 MA^2 omega^2`  .... (i)

And, average potential energy over one cycle is given as:

`(E_p)_'Avg" = 1/T int_0^T E_p dt`

`= 1/T int_0^T 1/2 Momega^2 A^2 sin^2 omegat dt`

`= 1/2T Momega^2 A^2 int_0^T ((1-cos 2 omegat))/2 dt`

`= 1/4T Momega^2A^2[(t - sin 2omegat)/2omega]_0^T`

`= 1/"4T" M omega^2 A^2 (T)`

`= (Momega^2A^2)/4`   .... (ii)

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.

Solution 2

 Let the particle executing SHM starts oscillating from its mean position. Then displacement equation is

`x= A sin omega t`

Particle velocity, `v = Aomega cos omegat`

Instantaneous K.E, `K = 1/2 mv^2 = 1/2 mA^2 omega^2 cos^2 omegat`

Average value of K.E over one complete cycle

`K_"av' = 1/T int_0^T mA^2omega^2 cos^2 omegat dt = (mA^2omega^2)/2T int_0^Tcos^2 omegat dt`

`= (mA^2omega^2)/(2T) int_0^T (1+cos 2omegat)/2 dt`

`= (mA^2omega^2)/(4T) [t +(sin 2 omegat)/(2omega)]_0^T`

`= (mA^2omega^2)/4T[(T-0)+ ((sin 2 omegat - sin 0)/(2omega))]`

`= 1/4 mA^2omega^2`  ... (i)

Again instantaneous P.E, `U = 1/2 kx^2 = 1/2 momega^2 x^2 = 1/2 momega^2 A^2 sin^2 omegat`

:. Average value of P.E over one complete cycle

`uu_"av" = 1/2 int_0^T 1/2 momega^2 A^2 sin^2 omegat = (mv^2 A^2)/(2T) int_0^T sin ^2 omegat dt`

`= (momega^2A^2)/(2T) int_0^T ((1-cos 2omegat))/2 dt`

`= (momega^2A^2)/4T [t -(sin 2omegat)/(2omega)]_0^T`

`= (momega^2A^2)/(4T) = [(T-0)- (sin 2omegat -sin 0)/(2omega)]`

`= 1/4 momega^2A^2`  .....(ii)

Simple comparision of  (i) and (ii) shows that

`K_"av" = uu_"av" = 1/4 momega^2A^2`

Concept: Force Law for Simple Harmonic Motion
  Is there an error in this question or solution?


NCERT Class 11 Physics
Chapter 14 Oscillations
Q 22 | Page 361
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