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

Topic: Physics - Simple Harmonic Motion (SHM)
Step 1: Understanding the Question:
In SHM, velocity varies with displacement. We are given the maximum velocity (at the mean position) and the amplitude. We need to find the displacement \(x\) for a specific velocity.
Step 2: Key Formula or Approach:
1. Maximum velocity: \(v_{\text{max}} = \omega A\).
2. Velocity at any displacement \(x\): \(v = \omega \sqrt{A^2 - x^2}\).
Step 3: Detailed Explanation:
1. Calculate the angular frequency \(\omega\):
Given \(v_{\text{max}} = 6 \, \text{cm/s}\) and \(A = 4 \, \text{cm}\).
\[ 6 = \omega \times 4 \implies \omega = 1.5 \, \text{rad/s} \]
2. Use the general velocity formula for \(v = 2 \, \text{cm/s}\):
\[ 2 = 1.5 \sqrt{4^2 - x^2} \]
\[ \frac{2}{1.5} = \sqrt{16 - x^2} \implies \frac{4}{3} = \sqrt{16 - x^2} \]
3. Square both sides to solve for \(x\):
\[ \frac{16}{9} = 16 - x^2 \]
\[ x^2 = 16 - \frac{16}{9} = 16 \left( 1 - \frac{1}{9} \right) = 16 \left( \frac{8}{9} \right) \]
\[ x = \sqrt{\frac{128}{9}} = \frac{\sqrt{64 \times 2}}{3} = \frac{8\sqrt{2}}{3} \, \text{cm} \]
Step 4: Final Answer:
The position is \(\frac{8\sqrt{2}}{3} \, \text{cm}\).