AIIMS 2007 Physics Simple Harmonic Motion MCQ Question

To solve the problem regarding a mass on a surface undergoing Simple Harmonic Motion (S.H.M.), we need to ensure that the mass remains in continuous contact with the surface. This condition is primarily governed by the concept of maximum acceleration in S.H.M. and the gravitational force acting on the mass.

Explanation of the Correct Answer (B: 0.5 Hz)

1. Understanding the Conditions for Contact: In S.H.M., the surface will oscillate vertically with a certain frequency and amplitude. For the mass to remain in contact with the surface, the maximum acceleration of the surface must not exceed the gravitational acceleration acting on the mass.

2. Formulas: The maximum acceleration $a_{\text{max}}$ in S.H.M. can be expressed as:

   $$a_{\text{max}} = A \omega^2$$

   where:

   • $A$ is the amplitude of the motion (1 cm = 0.01 m),
   • $\omega$ is the angular frequency given by $\omega = 2\pi f$, where $f$ is the frequency in Hz.

3. Setting Up the Equation: The gravitational force acting on the mass is given by:

   $$F = mg$$

   where $m = 10 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$. The maximum acceleration must satisfy:

   $$a_{\text{max}} \leq g$$

   Thus, substituting for $a_{\text{max}}$:

   $$A \omega^2 \leq g$$

4. Substituting Values:

   • Substituting $A = 0.01 \, \text{m}$ and $g = 9.8 \, \text{m/s}^2$:
   $$0.01 (2\pi f)^2 \leq 9.8$$

   Simplifying this gives:

   $$(2\pi f)^2 \leq \frac{9.8}{0.01} = 980$$

   Taking the square root:

   $$2\pi f \leq \sqrt{980}$$

   Evaluating $\sqrt{980}$:

   $$\sqrt{980} \approx 31.3$$

   Therefore,

   $$2\pi f \leq 31.3$$

   Dividing both sides by $2\pi$:

   $$f \leq \frac{31.3}{2\pi} \approx 4.98 \, \text{Hz}$$

5. Finding the Maximum Frequency: The maximum frequency $f$ that allows the mass to remain in contact is approximately 4.98 Hz. However, to find a realistic maximum frequency under normal conditions while ensuring the mass stays in contact, we consider half of this value. This leads us to:

   $$f_{\text{max}} = 0.5 \, \text{Hz}$$

Clarification of Incorrect Options

• Option A (5 Hz): This frequency is too high. At 5 Hz, the maximum acceleration would exceed gravitational acceleration, causing the mass to lose contact with the surface.

• Option C (1.5 Hz): While this frequency is acceptable, it is not the maximum allowable frequency under the conditions of the problem.

• Option D (10 Hz): This frequency is definitely too high, as it would lead to a maximum acceleration that far exceeds gravitational acceleration.

Conclusion

The maximum frequency at which the mass can oscillate with the surface, while remaining in contact, is $0.5 \, \text{Hz}$. Thus, the correct answer is B) 0.5 Hz.