Question

The variation of displacement with time of a simple harmonic motion (SHM) for a particle of mass \( m \) is represented by: \[ y = 2 \sin \left( \frac{\pi}{2} + \phi \right) \, \text{cm} \] The maximum acceleration of the particle is:

• A. \( \frac{\pi^2}{2} \, \text{cm/sec}^2 \)
• B. \( \frac{\pi}{2m} \, \text{cm/sec}^2 \)
• C. \( \frac{\pi^2}{2m} \, \text{cm/sec}^2 \)
• D. \( \frac{\pi^2}{2} \, \text{cm/sec}^2 \)

Hint:

In SHM, the maximum acceleration is given by \( a_{\text{max}} = \omega^2 A \), where \( \omega \) is the angular frequency and \( A \) is the amplitude.

Answer 1

The Correct Option is A

Step 1: Understanding the SHM Equation The displacement equation for SHM is given as: \[ y = A \sin (\omega t + \phi) \] Where: - \( y \) is the displacement, - \( A \) is the amplitude, - \( \omega \) is the angular frequency, - \( t \) is the time, - \( \phi \) is the phase constant. The given displacement equation is: \[ y = 2 \sin \left( \frac{\pi}{2} + \phi \right) \, \text{cm} \] Here, the amplitude \( A = 2 \) cm, and the angular frequency \( \omega \) is \( \frac{\pi}{2} \). Step 2: Formula for Maximum Acceleration The acceleration in SHM is given by: \[ a = -\omega^2 y \] The maximum acceleration occurs when \( y = A \), so: \[ a_{\text{max}} = \omega^2 A \] Step 3: Substituting the Known Values We know that: - \( A = 2 \, \text{cm} \) - \( \omega = \frac{\pi}{2} \) Now, substituting these values into the formula for maximum acceleration: \[ a_{\text{max}} = \left( \frac{\pi}{2} \right)^2 \times 2 \] Simplifying: \[ a_{\text{max}} = \frac{\pi^2}{4} \times 2 = \frac{\pi^2}{2} \, \text{cm/sec}^2 \] Step 4: Conclusion The maximum acceleration of the particle is \( \frac{\pi^2}{2} \, \text{cm/sec}^2 \), which corresponds to: \[ \boxed{(A)} \, \frac{\pi^2}{2} \, \text{cm/sec}^2 \]