# 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

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.

#### Solution 1

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

x =Asin omegat

Where,

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
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#### APPEARS IN

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