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A linear harmonic oscillator of force constant 2x10^6 N/m and amplitude of 0.01m has a total mechanical energy of 160J. Find maximum kinetic energy and maximum potential energy. [IIT 1989]

A linear harmonic oscillator of force constant 2x10^6 N/m and amplitude of 0.01m has a total mechanical energy of 160J. Find maximum kinetic energy and maximum potential energy. [IIT 1989]

Grade:11

2 Answers

Samridhi Bhatti
11 Points
8 years ago
We know that maximum potential energy=total mechanical energy. Thus, maximum P.E.=160J. For calculating the maximum kinetic energy, the formula used is max K.E.=1/2*k*a^2. (and not the usual 1/2*k(a^2-x^2) So, maximum K.E.=1/2*2x10^6*(0.01^2) which is equal to 100J.
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

As, we know total mechanical energy = maximum potential energy
∴ATQ.
T.E = 160J
⇒ Max.P.E.= 160J

Also, for maximum kinetic energy,
we know,
K.E.= 0.5​K(xm^2​)
wherexm ​= (0.01)m
& K=2×10^6Nm−1
⇒K.E.=(0.5​)(2×10^6)(0.01)^2
=100J

Thanks and Regards

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