Question

The relation between the force (F in newton) acting on a particle executing simple harmonic motion and the displacement of the particle (y in metre) is given by: \[ 500F + \pi^2 y = 0 \] If the mass of the particle is 2 g, the time period of oscillation of the particle is:

• A. \( 8 \text{ s} \)
• B. \( 6 \text{ s} \)
• C. \( 2 \text{ s} \)
• D. \( 4 \text{ s} \)

Hint:

In SHM, the time period depends on both the mass of the oscillating particle and the force constant. The formula \( T = 2\pi \sqrt{\frac{m}{k}} \) is crucial for solving such problems efficiently.

Answer 1

The Correct Option is C

Step 1: Identifying the SHM equation
The equation of motion for simple harmonic motion (SHM) is generally given as: \[ F = - k y \] where \( k \) is the force constant. From the given equation: \[ 500F + \pi^2 y = 0 \] Rearranging, \[ F = -\frac{\pi^2}{500} y \] Comparing with the standard SHM equation, we identify: \[ k = \frac{\pi^2}{500} \] Step 2: Using the SHM formula for time period
The time period of SHM is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] Given that the mass of the particle is \( m = 2 \) g = \( 2 \times 10^{-3} \) kg, we substitute: \[ T = 2\pi \sqrt{\frac{2 \times 10^{-3}}{\frac{\pi^2}{500}}} \] Step 3: Simplifying the expression
\[ T = 2\pi \sqrt{\frac{2 \times 10^{-3} \times 500}{\pi^2}} \] \[ T = 2\pi \sqrt{\frac{1}{\pi^2}} \] \[ T = 2\pi \times \frac{1}{\pi} \] \[ T = 2 \text{ s} \]