Question

A particle executes a linear SHM with an amplitude \(a\) and angular velocity \(\omega\). The ratio between its acceleration amplitude and displacement amplitude is

• A. \( \frac{\omega}{4} \)
• B. \( \omega^2 \)
• C. \( \omega \)
• D. \( \frac{\omega}{2} \)
• E. \( 2\omega \)

Hint:

In SHM, the amplitude of the acceleration is proportional to the square of the angular frequency, \( \omega^2 \).

Answer 1

The Correct Option is B

In simple harmonic motion (SHM), the amplitude of the acceleration \(A_a\) and displacement \(A_x\) are related by the following equations: 
- Displacement: \( x(t) = A_x \cos(\omega t) \) 
- Acceleration: \( a(t) = -A_a \omega^2 \cos(\omega t) \) 
From these equations, we can see that the acceleration amplitude \(A_a\) is related to the displacement amplitude \(A_x\) by: \[ A_a = A_x \omega^2 \] 
Therefore, the ratio of the acceleration amplitude to the displacement amplitude is: \[ \frac{A_a}{A_x} = \omega^2 \] 
Hence, the correct answer is (B).