Question

In simple harmonic motion (SHM), the acceleration of a particle at a displacement of 3m from the mean position is $48\text{ m/s}^2$ directed towards the mean position. Find the angular frequency.

Hint:

The negative sign in $a = -\omega^2 x$ simply indicates direction (towards the mean position). When doing purely numerical calculations with magnitudes, you can safely drop the negative sign.

Answer 1

Step 1: Understanding the Concept:
The defining characteristic of Simple Harmonic Motion (SHM) is that the acceleration of the oscillating particle is always strictly directly proportional to its displacement from the mean position.
Furthermore, this acceleration vector is always directed precisely opposite to the displacement vector (towards the equilibrium point).
Step 2: Key Formula or Approach:
The fundamental kinematic equation for acceleration $a$ in SHM at any given displacement $x$ is:
\[ a = -\omega^2 x \] Here, $\omega$ represents the angular frequency of the oscillator. Taking just the magnitude gives $|a| = \omega^2 x$.
Step 3: Detailed Explanation:
From the problem text, the specific magnitude of acceleration is explicitly given as $|a| = 48\text{ m/s}^2$.
The corresponding displacement magnitude from the mean position is explicitly $x = 3\text{ m}$.
Substitute these known values directly into the magnitude equation:
\[ 48 = \omega^2 \times 3 \] Rearrange the equation to isolate the unknown $\omega^2$:
\[ \omega^2 = \frac{48}{3} \] \[ \omega^2 = 16 \] Take the square root of both sides to confidently find the angular frequency:
\[ \omega = \sqrt{16} = 4\text{ rad/s} \] Step 4: Final Answer:
The angular frequency of the particle's motion is $4\text{ rad/s}$.