Question

A mass of 1 kg is attached to a spring of force constant 100 N/m. Time period is:

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

Hint:

The time period is independent of the amplitude of oscillation. Whether you pull the spring a little or a lot, as long as it stays within its elastic limit, the time taken for one full bounce remains the same.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement.
A classic example is the mass-spring system.
The time period of such a system is the time it takes to complete one full cycle of vibration (back and forth).
This period depends on the inertia of the system (mass) and the "stiffness" of the system (spring constant).
Higher mass leads to a slower oscillation (longer period) due to greater inertia.
A higher spring constant leads to a faster oscillation (shorter period) due to a stronger restoring force.
Note that the time period is independent of the gravitational constant \(g\) and the amplitude of the motion.
Step 2: Key Formula or Approach:
Standard formula for Time Period: \(T = 2\pi \sqrt{\frac{m}{k}}\).
Step 3: Detailed Explanation:
Identify the given parameters:
Mass \(m = 1\) kg.
Force constant \(k = 100\) N/m.
Substitute these values into the time period equation:
\[ T = 2\pi \sqrt{\frac{1}{100}} \]
Simplify the radical expression:
\[ \sqrt{\frac{1}{100}} = \frac{1}{10} \]
Therefore:
\[ T = 2\pi \cdot \frac{1}{10} = \frac{2\pi}{10} = 0.2\pi \text{ s} \]
The system takes approximately \(0.628\) seconds (\(0.2 \times 3.14\)) to complete one oscillation.
Step 4: Final Answer:
The time period of the oscillation is \(0.2\pi\) s.