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

Step 1: Understanding the Question:
We are given the amplitude, displacement from the mean position, and angular frequency of a particle in Simple Harmonic Motion (SHM). We need to calculate its acceleration at the given displacement.
Step 2: Key Formula or Approach:
The acceleration \( a \) of a particle in SHM is directly proportional to its displacement \( x \) from the mean position and is directed towards the mean position. The formula is:
\[ a = -\omega^2 x \] where \( \omega \) is the angular frequency. The question asks for the magnitude of the acceleration.
\[ |a| = \omega^2 |x| \] Step 3: Detailed Explanation:
Given values:
Amplitude \( A = 5 \) cm (This information is not needed for this specific calculation).
Displacement \( x = 3 \) cm.
Angular frequency \( \omega = 2 \) rad/s.
Substitute the values into the formula for the magnitude of acceleration:
\[ |a| = (2 \, \text{rad/s})^2 \times (3 \, \text{cm}) \] \[ |a| = 4 \, \text{s}^{-2} \times 3 \, \text{cm} \] \[ |a| = 12 \, \text{cm/s}^2 \] Step 4: Final Answer:
The acceleration of the particle is \( 12 \, \text{cm/s}^2 \).