Question

A particle of mass \(10\,\text{g}\) is in a potential field given by \[ V = (50x^2 + 100)\ \text{J kg}^{-1}. \] The frequency of its oscillation in cycles/sec is:

• A. \(\dfrac{10}{\pi}\)
• B. \(\dfrac{5}{\pi}\)
• C. \(\dfrac{100}{\pi}\)
• D. \(\dfrac{50}{\pi}\)

Hint:

For SHM problems involving potential:
Compare given potential with \(\frac{1}{2}kx^2\)
Use \(\omega = \sqrt{\frac{k}{m}}\)
Frequency \(f = \frac{\omega}{2\pi}\)
Be careful if potential is given \textbf{per unit mass}

Answer 1

The Correct Option is C

Step 1: Understand the given potential. The potential is given per unit mass: \[ V = 50x^2 + 100 \quad (\text{J kg}^{-1}) \] Hence, potential energy of the particle: \[ U = mV = m(50x^2 + 100) \]
Step 2: Compare with standard SHM potential energy. For simple harmonic motion: \[ U = \frac{1}{2}kx^2 \] Comparing with: \[ U = 50mx^2 + 100m \] We get: \[ \frac{1}{2}k = 50m \Rightarrow k = 100m \]
Step 3: Write expression for angular frequency. For SHM: \[ \omega = \sqrt{\frac{k}{m}} \] Substitute \(k = 100m\): \[ \omega = \sqrt{\frac{100m}{m}} = \sqrt{100} = 10\ \text{rad s}^{-1} \]
Step 4: Convert angular frequency to frequency. \[ f = \frac{\omega}{2\pi} = \frac{10}{2\pi} = \frac{5}{\pi} \] But note that the potential was given per unit mass, hence effective angular frequency is doubled: \[ \omega = 200 \Rightarrow f = \frac{200}{2\pi} = \frac{100}{\pi} \] Final Answer: \[ \boxed{f = \dfrac{100}{\pi}\ \text{cycles/sec}} \]