Question

A mass of \( 1 \text{ kg} \) is attached to a spring of force constant \( 100 \text{ N/m} \). Time period is:

• A. \( 0.2\pi \text{ s} \)
• B. \( 0.1\pi \text{ s} \)
• C. \( 0.5\pi \text{ s} \)
• D. \( 2\pi \text{ s} \)

Hint:

When evaluating time periods for spring-mass systems, remember that the time period is independent of the amplitude of oscillation or gravitational acceleration \( g \). It depends solely on the ratio of mass to spring stiffness (\(m/k\))!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A mass attached to a spring executes Simple Harmonic Motion (SHM) when displaced from its equilibrium position.
The time period \(T\) of this oscillation is determined by the balance between the inertia of the mass \(m\) and the stiffness of the spring, represented by the force constant \(k\).
A higher mass increases the time period due to inertia, while a higher spring constant decreases the time period due to stronger restoring forces.
Key Formula or Approach:
The time period \(T\) for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where:
\(m = \text{mass in kg}\)
\(k = \text{spring constant in N/m}\)
Step 2: Detailed Explanation:
From the question, we identify the given physical parameters:
\(m = 1 \text{ kg}\)
\(k = 100 \text{ N/m}\)
Substitute these values into the standard formula for the time period:
\[ T = 2\pi \sqrt{\frac{1}{100}} \]
The square root of \(1/100\) is \(1/10\), or \(0.1\).
\[ T = 2\pi \cdot 0.1 = 0.2\pi \text{ s} \]
This calculated value represents the time taken for the mass to complete one full cycle of oscillation.
Step 3: Final Answer:
The time period of the mass-spring system is \(0.2\pi \text{ s}\), which corresponds to option (A).