Step 1: Understanding the Concept:
A mass suspended on a vertical spring oscillates in Simple Harmonic Motion (S.H.M.) about its equilibrium position.
The equilibrium position is where the spring force balances the gravitational force.
Step 2: Key Formulas or Approach:
Time period of a spring-mass system: \( T = 2\pi\sqrt{\frac{m}{k}} \).
Angular frequency: \( \omega = \frac{2\pi}{T} = \sqrt{\frac{k}{m}} \).
Maximum speed in S.H.M.: \( v_{\text{max}} = A\omega \), where \( A \) is the amplitude.
Equilibrium extension: \( x_0 = \frac{mg}{k} \).
Step 3: Detailed Explanation:
Given the time period \( T = 0.1\text{ s} \).
The angular frequency is \( \omega = \frac{2\pi}{T} = \frac{2\pi}{0.1} = 20\pi\text{ rad/s} \).
We also know that \( \sqrt{\frac{k}{m}} = \omega \), so \( \frac{m}{k} = \frac{1}{\omega^2} \).
The problem states that the spring is unstretched at the highest point of its oscillation.
This implies the highest point is the natural length of the spring.
The system oscillates symmetrically about its equilibrium position.
The equilibrium position is lower than the natural length by an extension \( x_0 \).
At equilibrium, the net force is zero: \( kx_0 = mg \), so \( x_0 = \frac{mg}{k} \).
Since the highest point is the unstretched position, the distance from equilibrium to the highest point is the amplitude \( A \).
Therefore, the amplitude is \( A = x_0 = \frac{mg}{k} \).
We can rewrite the amplitude in terms of \( \omega \):
\[ A = g \left(\frac{m}{k}\right) = g \left(\frac{1}{\omega^2}\right) = \frac{g}{\omega^2} \]
Now, calculate the maximum speed:
\[ v_{\text{max}} = A\omega = \left(\frac{g}{\omega^2}\right) \omega = \frac{g}{\omega} \]
Substitute the known values \( g = 10\text{ m/s}^2 \) and \( \omega = 20\pi\text{ rad/s} \):
\[ v_{\text{max}} = \frac{10}{20\pi} = \frac{1}{2\pi}\text{ m/s} \]
Step 4: Final Answer:
The maximum speed of the mass is \( \frac{1}{2\pi}\text{ m/s} \).