Question

The variation of displacement with time of a simple harmonic motion is \[ y = 2\sin\left(\frac{\pi t}{2}+\phi\right)\text{ cm}. \] The maximum acceleration of the particle is

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

Hint:

SHM formulas: \begin{itemize} \item \( a_{\max} = \omega^2 A \). \end{itemize}

Answer 1

The Correct Option is D

Concept: In SHM: \[ a_{\max} = \omega^2 A \] Step 1: {\color{red}Identify parameters.} Amplitude: \[ A = 2\,\text{cm} \] Angular frequency: \[ \omega = \frac{\pi}{2} \] Step 2: {\color{red}Maximum acceleration.} \[ a_{\max} = \left(\frac{\pi}{2}\right)^2 \times 2 = \frac{\pi^2}{2} \]