Question

The amplitude of a simple harmonic oscillator is \(A\). When the velocity of particle is half of its maximum velocity, then its position is at

• A. \(\frac{A}{2}\)
• B. \(\frac{\sqrt3A}{4}\)
• C. \(\frac{A}{4}\)
• D. \(\frac{\sqrt3A}{2}\)

Hint:

In SHM, use \(v=\omega\sqrt{A^2-x^2}\) to relate velocity and displacement.

Answer 1

The Correct Option is D

Concept:
In SHM, velocity at displacement \(x\) is: \[ v=\omega\sqrt{A^2-x^2} \] and maximum velocity is: \[ v_{\max}=\omega A \]

Step 1: Given: \[ v=\frac{v_{\max}}{2} \]

Step 2: Substitute: \[ \omega\sqrt{A^2-x^2}=\frac{\omega A}{2} \]

Step 3: Cancel \(\omega\): \[ \sqrt{A^2-x^2}=\frac{A}{2} \]

Step 4: Square both sides: \[ A^2-x^2=\frac{A^2}{4} \] \[ x^2=A^2-\frac{A^2}{4} \] \[ x^2=\frac{3A^2}{4} \]

Step 5: \[ x=\frac{\sqrt3A}{2} \] \[ \boxed{\frac{\sqrt3A}{2}} \]