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: SHM Equation The displacement equation for SHM is: \[\ny = A \sin (\omega t + \phi)\n\] Where: - \( y \) is displacement, - \( A \) is amplitude, - \( \omega \) is angular frequency, - \( t \) is time, - \( \phi \) is phase constant. Given displacement equation: \[\ny = 2 \sin \left( \frac{\pi}{2} + \phi \right) \, \text{cm}\n\] Here, \( A = 2 \) cm, and \( \omega \) is \( \frac{\pi}{2} \). Step 2: Maximum Acceleration Formula Acceleration in SHM is: \[\na = -\omega^2 y\n\] Maximum acceleration occurs when \( y = A \): \[\na_{\text{max}} = \omega^2 A\n\] Step 3: Value Substitution Known values: - \( A = 2 \, \text{cm} \), - \( \omega = \frac{\pi}{2} \). Substitute into the maximum acceleration formula: \[\na_{\text{max}} = \left( \frac{\pi}{2} \right)^2 \times 2\n\] Simplifying: \[\na_{\text{max}} = \frac{\pi^2}{4} \times 2 = \frac{\pi^2}{2} \, \text{cm/sec}^2\n\] Step 4: Conclusion Maximum acceleration is \( \frac{\pi^2}{2} \, \text{cm/sec}^2 \), which is: \[\n\boxed{(A)} \, \frac{\pi^2}{2} \, \text{cm/sec}^2\n\]