Question

The potential energy of a particle of mass 10 g as a function of displacement \( x \) is \( (50 \, x^2 + 100) \). The frequency of oscillation is:

• A. \( \frac{10}{\pi} \) s\(^{-1}\)
• B. \( \frac{5}{\pi} \) s\(^{-1}\)
• C. \( \frac{100}{\pi} \) s\(^{-1}\)
• D. \( \frac{50}{\pi} \) s\(^{-1}\)

Hint:

For a harmonic oscillator, the frequency is given by \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). Always compare the given potential energy function with \( U = \frac{1}{2} k x^2 \) to determine \( k \).

Answer 1

The Correct Option is D

Step 1: Identifying the force constant
The potential energy of a simple harmonic oscillator is given by: \[ U = \frac{1}{2} k x^2. \] Comparing with the given function: \[ U = 50 x^2 + 100, \] we identify the force constant: \[ \frac{1}{2} k = 50 \quad \Rightarrow \quad k = 100. \] Step 2: Calculating angular frequency
The angular frequency \( \omega \) is given by: \[ \omega = \sqrt{\frac{k}{m}}. \] Given: - \( k = 100 \), - \( m = 10 \) g = \( 0.01 \) kg. \[ \omega = \sqrt{\frac{100}{0.01}}. \] \[ \omega = \sqrt{10000} = 100 { rad/s}. \] Step 3: Finding frequency of oscillation
The frequency of oscillation is: \[ f = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi}. \] Step 4: Conclusion
Thus, the frequency of oscillation is: \[ \frac{50}{\pi} { s}^{-1}. \]

Answer 2

The potential energy of a particle is given by the function \( U(x) = 50x^2 + 100 \). To find the frequency of oscillation, we need to treat this as a simple harmonic motion and use the formula for potential energy in simple harmonic motion, \( U(x) = \frac{1}{2} k x^2 \), where \( k \) is the force constant. Comparing the given potential energy \( 50x^2 + 100 \) with \( \frac{1}{2} k x^2 \), it follows that:

\( \frac{1}{2} k = 50 \)

This implies:

\( k = 100 \, \text{N/m} \)

The mass of the particle \( m \) is \( 10 \, \text{g} = 0.01 \, \text{kg} \). The formula for the angular frequency \( \omega \) in simple harmonic motion is:

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

Substituting the values:

\( \omega = \sqrt{\frac{100}{0.01}} = \sqrt{10000} = 100 \) rad/s

The frequency \( f \) is given by:

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

Therefore,

\( f = \frac{100}{2\pi} = \frac{50}{\pi} \) s\(^{-1}\)

Thus, the frequency of oscillation is \(\frac{50}{\pi}\) s\(^{-1}\).