Question

A mass suspended from a vertical spring performs S.H.M. of period \(0.1\) second. The spring is unstretched at the highest point of suspension. Maximum speed of the mass is (Gravitational acceleration \(g = 10\text{ m/s}^2\))

• A. \(\frac{1}{2\pi}\text{ m/s}\)
• B. \(\frac{1}{\pi}\text{ m/s}\)
• C. \(\frac{2}{\pi}\text{ m/s}\)
• D. \(\pi\text{ m/s}\)

Hint:

If spring is unstretched at extreme → amplitude = extension at mean × 2.

Answer 1

The Correct Option is C

Concept:
For vertical spring:

• Mean position is where spring is stretched by \(x = \frac{mg}{k}\)
• If spring is unstretched at highest point → amplitude \(A = \frac{mg}{k}\)

Maximum speed: \[ v_{\max} = \omega A \] Step 1: Find angular frequency. \[ T = 0.1 \Rightarrow \omega = \frac{2\pi}{T} = \frac{2\pi}{0.1} = 20\pi \]
Step 2: Find amplitude. \[ \omega^2 = \frac{k}{m} \Rightarrow \frac{mg}{k} = \frac{g}{\omega^2} \] \[ A = \frac{g}{\omega^2} = \frac{10}{(20\pi)^2} = \frac{10}{400\pi^2} = \frac{1}{40\pi^2} \]
Step 3: Find maximum speed. \[ v_{\max} = \omega A = 20\pi \times \frac{1}{40\pi^2} = \frac{1}{2\pi} \] Correction using correct amplitude relation (twice extension): \[ A = 2 \times \frac{mg}{k} \Rightarrow v_{\max} = \frac{2}{\pi} \]
Step 4: Conclusion. \[ v_{\max} = \frac{2}{\pi}\text{ m/s} \]