Question

A mass is suspended from a vertical spring which is executing S.H.M. of frequency \(10\) Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is (\(g=10\text{m/s}^{2}\))

• A. \(2\pi\) m/s
• B. \(\pi\) m/s
• C. \(\left(\frac{1}{2\pi }\right)\) m/s
• D. \(\frac{1}{4\pi }m/s\)

Hint:

The amplitude equals the equilibrium extension \(g/\omega^{2}\) because the spring is unstretched at the top.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The mass oscillates about the equilibrium position, where the spring is stretched by \(x_{0}=\dfrac{mg}{k}\). At the highest point the spring is unstretched, so the amplitude is \(A=x_{0}\).

Step 2: Angular Frequency
\(\omega=2\pi f=20\pi\) rad/s
Since \(\omega^{2}=\dfrac{k}{m}\), \(x_{0}=\dfrac{g}{\omega^{2}}=\dfrac{10}{400\pi^{2}}\) m

Step 3: Maximum Speed
\(v_{max}=A\omega=\dfrac{g}{\omega}=\dfrac{10}{20\pi}\)

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

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