Question

A particle performs linear S.H.M. with potential energy \(U = 0.1\pi^2 x^2\). If the mass is \(20\,g\), what is its frequency?

• A. \(1.581\,\text{Hz}\)
• B. \(3.162\,\text{Hz}\)
• C. \(0.790\,\text{Hz}\)
• D. \(6.283\,\text{Hz}\)

Hint:

In simple harmonic motion, if the potential energy is given in the form \(ax^2\), compare it with \( \frac{1}{2}kx^2 \) to find the force constant \(k\). Once \(k\) is known, use \( \omega = \sqrt{k/m} \) and \( f=\frac{\omega}{2\pi} \) to quickly determine the frequency.

Answer 1

The Correct Option is A

Concept: For a particle executing simple harmonic motion, the potential energy is expressed as \[ U = \frac{1}{2}kx^2 \] where \(k\) is the force constant of the system. By comparing the given potential energy expression with this standard form, the value of the force constant can be determined. The angular frequency of oscillation is given by \[ \omega = \sqrt{\frac{k}{m}} \] and the frequency of oscillation is \[ f = \frac{\omega}{2\pi} \] These relations connect the restoring force constant, mass of the particle, and frequency of oscillation.

Step 1: Compare the given potential energy with the standard form.
The given expression is \[ U = 0.1\pi^2 x^2 \] The standard expression is \[ U = \frac{1}{2}kx^2 \] Comparing the coefficients of \(x^2\): \[ \frac{1}{2}k = 0.1\pi^2 \] \[ k = 0.2\pi^2 \]

Step 2: Convert the mass into SI units.
The mass of the particle is given as \[ m = 20\,g = 0.02\,kg \]

Step 3: Calculate the angular frequency.
Using \[ \omega = \sqrt{\frac{k}{m}} \] \[ \omega = \sqrt{\frac{0.2\pi^2}{0.02}} \] \[ \omega = \sqrt{10\pi^2} \] \[ \omega = \pi\sqrt{10} \]

Step 4: Determine the frequency.
\[ f = \frac{\omega}{2\pi} \] \[ f = \frac{\pi\sqrt{10}}{2\pi} \] \[ f = \frac{\sqrt{10}}{2} \] \[ f \approx 1.581 \, \text{Hz} \]

Step 5: Final result.
Thus, the frequency of oscillation of the particle is \[ \boxed{1.581\,\text{Hz}} \]