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Chapter 13: Problem 69

A particle of mass \(m\) has potential energy given by \(U=a x^{2}\) where \(a\) is a constant and \(x\) is the particle's position. Find an expression for the frequency of simple harmonic oscillations this particle undergoes.

Short Answer

Expert verified
The frequency of the simple harmonic motion is \( \frac{1}{2\pi} \sqrt{\frac{2a}{m}} \).

Step by step solution

01

Express Potential Energy in terms of Force

The force acting on a particle in a potential field is given by the negative derivative of the potential energy with respect to position. Therefore, we can write: \( F = -\dfrac{dU}{dx} \). Substitute \( U = a x^{2} \) into this equation.
02

Derive the Force Equation

Differentiating \( ax^{2} \) with respect to \( x \), the force acting on the particle is obtained: \( F = -2ax \).
03

Compare with the force in Hooke's Law

The restoring force in a simple harmonic motion scenario is given by Hooke's law, which states: \( F = -kx \). By equating this to the obtained force, \( -kx = -2ax \). It can be seen that the force constant \( k = 2a \).
04

Find the frequency

The frequency \( f \) is given by \( \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} \), substituting \( k = 2a \): \( f = \dfrac{1}{2\pi} \sqrt{\dfrac{2a}{m}} \).

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