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.

Suppose a Body of mass m is undergoing Simple harmonic motion i.e. oscillating to and fro from a fixed position called mean Position with time period T of one oscillation, and the maximum displacement of the body from the mean position is called the amplitude A of the body, now the position of the particle or displacement of particle from mean position at any instant of time is given by the relation

x = A sin ωt

where x is denotes the position of particle or displacement from mean position at any instant of time t, ω is the angular frequency of the Simple harmonic motion of the block, A is the amplitude

differentiating the equation resenting position of particle with respect to time to find velocity of particle at any instant

We know where v is the velocity of the particle

And differentiation where a and b are constants

So differentiating we get velocity of the body as

v = Aω cos ωt

we know kinetic energy of a particle is given by the relation

K = 1/2 mv^{2}

Where K is the kinetic energy of a particle of mass m and is moving with a velocity v

So instantaneous kinetic energy of the body undergoing Simple harmonic motion is

K = 1/2 m (A Aω cos ωt)^{2}

Or Instantaneous kinetic energy of the body is

K = 1/2 mA^{2}ω^{2}cos^{2}ωt

Now to find average kinetic Energy of the body in time interval T (time period of one complete oscillation) we will integrate the instantaneous kinetic energy with respect to time in the interval and then divide it with total time T

So average kinetic energy of the body denoted by K_{avg} will be

Since m,A, ω are constants we have

Solving further

We know relation between time period T and angular frequency ω ω = 2π/T

So we get

So the average kinetic energy of the body in one oscillation is

K_{avg} = 1/4 mA^{2}ω^{2}

Now the potential energy of a body undergoing Simple harmonic motion at any instant is given by the relation

U = 1/2 kx^{2}

Where k is the force constant of the spring, and x is the displacement of the body from mean position

We know the relation between force constant and angular frequency as

k = m ω^{2}

where k is the force constant, m is the mass of the body and v ω is the angular frequency of single harmonic motion

so the instantaneous potential energy of the body becomes

U = 1/2 mω^{2}x^{2}

And we know displacement of particle from mean position is

x = A sin ωt

where x is denotes the position of particle or displacement from mean position at any instant of time t, ω is the angular frequency of the Simple harmonic motion of the block, A is the amplitude

so putting the value of x we get, instantaneous Potential energy of the Body

U = 1/2 mA^{2}ω^{2}sin^{2}ωt

Now to find average Potential Energy of the body in time interval T (time period of one complete oscillation) we will integrate the instantaneous Potential energy with respect to time in the interval and then divide it with total time T

So average Potential energy of the body denoted by U_{avg} will be

Since m,A, ω are constants we have

Solving further

We know relation between time period T and angular frequency ω ω = 2π/T

So we get

So the average Potential energy of the body in one oscillation is

U_{avg} = 1/4 mA^{2}ω^{2}

So we get that Average Potential Energy and Average Kinetic Energy of the Body undergoing Simple Harmonic Oscillation in one cycle or oscillation are equal and

K_{avg} = U_{avg} = 1/4 mA^{2}ω^{2}

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