Question

A particle performs SHM with an amplitude of 5 cm. Find its acceleration when it is 3 cm from the mean position (assume \( \omega = 2 \, \text{rad/s} \)).

• A. \(3\ \text{cm/s}^2\)
• B. \(6\ \text{cm/s}^2\)
• C. \(12\ \text{cm/s}^2\)
• D. \(24\ \text{cm/s}^2\)

Hint:

In SHM, acceleration depends only on displacement from the mean position and angular frequency: \( a = -\omega^2 x \). The negative sign always indicates that the acceleration acts toward the mean position.

Answer 1

The Correct Option is C

Concept: In Simple Harmonic Motion (SHM), acceleration is directly proportional to displacement from the mean position and always directed toward the mean position. The relation is \[ a = -\omega^2 x \] where \(a\) = acceleration, \(\omega\) = angular frequency, \(x\) = displacement from mean position.
Step 1: {Substitute the given values.} \[ \omega = 2 \ \text{rad/s}, \qquad x = 3 \ \text{cm} \] \[ a = -\omega^2 x \] \[ a = -(2)^2 \times 3 \] \[ a = -4 \times 3 \] \[ a = -12 \ \text{cm/s}^2 \]
Step 2: {Interpret the result.} The negative sign indicates the acceleration is directed toward the mean position. Thus the magnitude of acceleration is \[ |a| = 12 \ \text{cm/s}^2 \]