#### Question

Obtain an expression for potential energy of a particle performing simple harmonic motion. Hence evaluate the potential energy

- at mean position and
- at extreme position.

#### Solution

Consider a particle of mass ‘m’ performing simple harmonic motion along the path AB about its mean position O as shown.

If the particle is at a distance x (Q) from its mean position O, then the restoring force F on the particle is

F = −kx (k = Force constant)

If the particle undergoes a further infinitesimal displacement dx against F, the work done is

dW= -Fdx

=- (-kx)dx

=kxdx

The total work done from O (x = 0) to Q (x = x) is

`intdW=int_0^x`

On integrating we get :

W=1/2kx^{2}

W=1/2mω^{2}x^{2} (∵ k = mω^{2})

This gives the potential energy of a particle executing simple harmonic motion.

i.e

Potential energy = 1/2kx^{2} = 1/2mω^{2}x^{2}

**At the mean position O:-** The velocity of the particle performing simple harmonic motion is maximum and the displacement is minimum (x = 0).

As x = 0

Potential energy = 1/2kx^{2} = 0

**At the extreme position A/B:-** The velocity of the particle performing simple harmonic motion is minimum and the displacement is maximum (x = ±a).

∴ Potential energy = 1/2kx^{2} = 1/2ka^{2} = 1/2mω^{2}a^{2}

**To calculate new frequency of rotation of disc:-**

It is given that

n_{1} = 100 r.p.m

The moment of inertia of the disc about transverse axis is I_{1} = 2 x 10^{-4} kg m^{2}

The mass of the wax is m = 20 g = 20 x 10^{-3} kg

The distance from the centre where wax falls = r = 5 cm = 5 x 10^{-2} m

To find n_{2} = ?

Now, the moment of inertia of the disc with wax on it is I_{2} = I_{1} + I_{wax} (according to parallel axes theorem)

So, we have I_{2} = I_{1} + mr^{2}

Now, we know that the angular momentum is conserved. So, we get

I_{1}ω_{1} = I_{2}ω_{2}

I_{1}(2πn_{1}) = (I_{1} + mr^{2}) (2πn_{2})

I_{1}n_{1} = (I_{1} + mr^{2})n_{2}

`thereforen_2=(I_1n_1)/(I_1+mr^2)=(2xx10^-4 xx100)/((2xx10^-4)+[20xx10^-3xx(5xx10^-2)^2])`

`thereforen_2=0.02/(2xx10^-4 + 0.5 xx 10^-4)=0.02/(2.5xx10^-4)=80`

Hence, the new frequency of rotation of disc is 80 revolutions per minute.