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Solution - Obtain an Expression for Potential Energy of a Particle Performing Simple Harmonic Motion. Hence Evaluate the Potential Energy - K.E.(Kinetic Energy) and P.E.(Potential Energy) in S.H.M.

Question

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

  1. at mean position and
  2. 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=1/2kx2

W=1/2mω2x2                 (∵ k = mω2)

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

i.e

Potential energy = 1/2kx2 = 1/2mω2x2

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 = 1/2kx2 = 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 = 1/2kx2 = 1/2ka2 = 1/2mω2a2

To calculate new frequency of rotation of disc:-

It is given that

n1 = 100 r.p.m

The moment of inertia of the disc about transverse axis is I1 = 2 x 10-4 kg m2

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 n2 = ?

Now, the moment of inertia of the disc with wax on it is I2 = I1 + Iwax (according to parallel axes theorem)

So, we have I2 = I1 + mr2

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

I1ω1 = I2ω2

I1(2πn1) = (I1 + mr2) (2πn2)

I1n1 = (I1 + mr2)n2

`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.

Is there an error in this question or solution?

APPEARS IN

2014-2015 (March)
Question 4.2 | 7 marks

Video TutorialsVIEW ALL [1]

Solution for question: Obtain an Expression for Potential Energy of a Particle Performing Simple Harmonic Motion. Hence Evaluate the Potential Energy concept: K.E.(Kinetic Energy) and P.E.(Potential Energy) in S.H.M.. For the courses HSC Science (Computer Science), HSC Science (Electronics), HSC Science (General)
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