Question

A particle in SHM has a speed of \(6\,cm/s\) at the mean position and an amplitude of \(4\,cm\). Find its position when its velocity is \(2\,cm/s\).

• A. \( \frac{8\sqrt{2}}{3} \,cm \)
• B. \( \frac{4\sqrt{2}}{3} \,cm \)
• C. \( \frac{8}{3} \,cm \)
• D. \( 2\sqrt{2} \,cm \)

Hint:

Important SHM relations:
• \(v_{max} = \omega A\)
• \(v = \omega\sqrt{A^2-x^2}\) Velocity is maximum at the mean position and zero at the extreme positions.

Answer 1

The Correct Option is A

Concept: For a particle performing Simple Harmonic Motion (SHM), the velocity at any position \(x\) is given by \[ v = \omega \sqrt{A^2 - x^2} \] where
• \(A\) = amplitude
• \(x\) = displacement from mean position
• \(\omega\) = angular frequency At the mean position, velocity is maximum: \[ v_{max} = \omega A \]

Step 1: Find angular frequency. Given \[ v_{max} = 6\,cm/s \] \[ A = 4\,cm \] Using \[ v_{max} = \omega A \] \[ 6 = \omega \times 4 \] \[ \omega = \frac{3}{2} \]

Step 2: Use the SHM velocity formula. \[ v = \omega \sqrt{A^2 - x^2} \] Substitute \(v=2\): \[ 2 = \frac{3}{2}\sqrt{16 - x^2} \]

Step 3: Solve the equation. \[ \sqrt{16-x^2} = \frac{4}{3} \] Square both sides: \[ 16 - x^2 = \frac{16}{9} \] \[ x^2 = \frac{128}{9} \] \[ x = \frac{8\sqrt2}{3} \] Thus, the position of the particle is \[ \boxed{\frac{8\sqrt2}{3}\,cm} \]