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:

Total mechanical energy in SHM stays constant, split between kinetic and potential energy at every position. Try writing this conservation equation between the mean position, where speed is maximum, and the point where the given velocity occurs — it lets you find the position without separately calculating the angular frequency.

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 \(\frac{8\sqrt{2}}{3} \, \text{cm}\).