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.
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
The total work done from O (x = 0) to Q (x = x) is
On integrating we get :
W=1/2mω2x2 (∵ k = mω2)
This gives the potential energy of a particle executing simple harmonic motion.
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=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.
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