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:

The mean position lies a distance mg over k below the unstretched position, and this distance is the amplitude; then the maximum speed is amplitude times angular frequency.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The spring is unstretched at the highest point, which is an extreme position. The equilibrium position is at extension \(x_0=\frac{mg}{k}\), so the amplitude is \(A=x_0\).

Step 2: Angular frequency
\(\omega=\frac{2\pi}{T}=\frac{2\pi}{0.1}=20\pi\text{ rad/s}\). Also \(\omega^2=\frac{k}{m}\), so \(x_0=\frac{g}{\omega^2}\).

Step 3: Maximum speed
\(v_{max}=A\omega=\frac{g}{\omega}=\frac{10}{20\pi}\).

Step 4: Compare the options
\(v_{max}=\frac{1}{2\pi}\text{ m/s}\), the first option.

Final Answer:
The maximum speed is \(\frac{1}{2\pi}\text{ m/s}\). \[ \boxed{\text{(A) }\dfrac{1}{2\pi}\text{ m/s}} \]