Question

A particle executes SHM with angular frequency \( \omega=0.5\,\text{rad s}^{-1} \) and amplitude \(A=5\,\text{cm}\). Find the acceleration when its displacement is \(x=4\,\text{cm}\).

• A. \(0.02\,\text{m/s}^2\)
• B. \(0.0125\,\text{m/s}^2\)
• C. \(0.01\,\text{m/s}^2\)
• D. \(0.005\,\text{m/s}^2\)

Hint:

In SHM, \[ a=-\omega^2x \] Acceleration depends only on displacement and angular frequency.

Answer 1

The Correct Option is C

Concept: In Simple Harmonic Motion, acceleration is proportional to displacement and is directed towards the mean position. \[ a=-\omega^2x \] The negative sign indicates direction only.

Step 1: Convert displacement into SI unit. \[ x=4\,\text{cm}=0.04\,\text{m} \] \[ \omega=0.5\,\text{rad/s} \]

Step 2: Apply SHM acceleration formula. \[ a=\omega^2x \] \[ a=(0.5)^2(0.04) \] \[ a=0.25\times0.04 \] \[ a=0.01\,\text{m/s}^2 \] Thus, \[ \boxed{a=0.01\,\text{m/s}^2} \] Hence the correct answer is \[ \boxed{(C)} \]