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Simple harmonic motion

Question

A particle is moving on x-axis has potential energy U=220x+5x2J along x-axis. The particle is released at x = - 3. The maximum value of 'x' is _________.

[x is in meters and U is in joule]

Moderate
Solution

Given that U=220x+5x2

Interaction force on particle is given as

F=dUdx=2010x

As F is linearly varying with -x, particle is executing SHM
At equilibrium position F = 0

 2010x=0 x=2

Since particle is released at x = - 3, therefore amplitude of particle is 5.

It will oscillate about x = 2 with an amplitude of 5.
 Maximum value of x will be 7.



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