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Question

The potential energy of a particle oscillating along X axis is given as U=15+4x+(2x3)2, where U is in Joules and x in metres. Total mechanical energy of the particle is 36 J. Mass of particle is 2 kg. Then,

A
Angular frequency of SHM will be 2 rad/sec
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B
Kinetic Energy at mean position =16 J
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C
Amplitude of SHM is 2 m
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D
None of these
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Solution

The correct options are
A Angular frequency of SHM will be 2 rad/sec
B Kinetic Energy at mean position =16 J
C Amplitude of SHM is 2 m
Given that
U=15+4x+(2x3)2
Since, the potential energy is only a function of x, for conservative forces,

F=dUdx
F=ddx(15+4x+(2x3)2)=(0+4+2(2x3)×2)=8x+8
F=8(x1)

Comparing this with F=k(xx0) we get, k=8 N/m

Given,
Mass of the particle (m)=2kg
So, the Angular frequency of the particle ω=km=82=2 rad/s

At mean position , F=0
8(x1)=0x=1 m


Potential energy at mean position (i.e) x=1 m
Umean=15+4x+(2x3)2=15+(4)+(2×13)2=20 J
Kinetic energy at mean position is given by
Kmean=EUmean=3620=16 J
We know that, kinetic energy at mean position is given by
Kmean=12mω2A2
16=12mω2A2

From the given data,
we can write that,
16=12×2×(2)2×A2
A2=4A=2 m
Thus, options (a) , (b) and (c) are the correct answers.

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