Question

The equation of motion of a particle executing simple harmonic motion is given by: \[ 4\frac{d^2y}{dt^2} + \pi^2 y = 0 \] where \( y \) is in meters and \( t \) is in seconds. The time period of oscillation of the particle is:

• A. \( 1 \, \text{s} \)
• B. \( 2 \, \text{s} \)
• C. \( 3 \, \text{s} \)
• D. \( 4 \, \text{s} \)

Hint:

For SHM, the time period is related to the angular frequency by \( T = \frac{2\pi}{\omega} \), where \( \omega \) is the angular frequency. In this case, \( \omega = \frac{\pi}{2} \), so the time period is 4 seconds.

Answer 1

The Correct Option is D

Step 1: Understand the equation of motion for SHM. The standard form of the equation of motion for SHM is: \[ \frac{d^2y}{dt^2} + \omega^2 y = 0 \] By comparing it with the given equation, we get: \[ 4\frac{d^2y}{dt^2} + \pi^2 y = 0 \quad \Rightarrow \quad \omega^2 = \frac{\pi^2}{4} \quad \Rightarrow \quad \omega = \frac{\pi}{2} \]
Step 2: Calculate the time period of oscillation. The time period \( T \) is given by: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{\pi}{2}} = 4 \, \text{seconds}. \]