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

To find the frequency of the particle performing linear Simple Harmonic Motion (SHM), we begin by understanding that the potential energy \( U \) is given by:

\(U = 0.1\pi^2 x^2\)

In SHM, the potential energy of the system can be expressed as:

\(U = \frac{1}{2} k x^2\)

Here, comparing both expressions, we can identify the spring constant \( k \) as:

\(\frac{1}{2}k = 0.1\pi^2 \Rightarrow k = 0.2\pi^2\)

Next, the angular frequency \( \omega \) of the SHM is related to the spring constant \( k \) and mass \( m \) by:

\(\omega = \sqrt{\frac{k}{m}}\)

The mass \( m \) is given as \( 20\, \text{g} = 0.02\, \text{kg} \). Substituting the values of \( k \) and \( m \) into the expression for \( \omega \) gives:

\(\omega = \sqrt{\frac{0.2\pi^2}{0.02}}\)

Simplifying the expression:

\(\omega = \sqrt{10\pi^2}\)

\(\omega = \pi\sqrt{10}\)

The frequency \( f \) is related to the angular frequency \( \omega \) by the formula:

\(f = \frac{\omega}{2\pi}\)

Substituting the calculated angular frequency:

\(f = \frac{\pi\sqrt{10}}{2\pi} = \frac{\sqrt{10}}{2}\)

\(f = \frac{3.162}{2} = 1.581\,\text{Hz}\)

Hence, the frequency of the particle's motion is \( 1.581\,\text{Hz} \), which matches the correct option. The other options can be ruled out as they do not satisfy this computation.