Question:
A mass of \( 1 \text{ kg} \) is attached to a spring of force constant \( 100 \text{ N/m} \). Time period is:
A mass of \( 1 \text{ kg} \) is attached to a spring of force constant \( 100 \text{ N/m} \). Time period is:
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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 (\(m/k\))!
Updated On: Jun 3, 2026
- \( 0.2\pi \text{ s} \)
- \( 0.1\pi \text{ s} \)
- \( 0.5\pi \text{ s} \)
- \( 2\pi \text{ s} \)
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The Correct Option is A
Solution and Explanation
Concept:
For a simple harmonic oscillator consisting of a mass \( m \) attached to an ideal elastic spring with force constant (or spring constant) \( k \), the time period of oscillation \( T \) is governed by the inertia of the mass and the stiffness of the spring.
The standard formula connecting these physical quantities is:
\[
T = 2\pi \sqrt{\frac{m}{k}}
\]
where:
• \( T \) is the time period in seconds (\(\text{s}\))
• \( m \) is the mass in kilograms (\(\text{kg}\))
• \( k \) is the spring constant in Newtons per meter (\(\text{N/m}\))
Step 1: Identifying the given parameters from the problem statement.
From the question, we are given:
• Mass attached, \( m = 1 \text{ kg} \)
• Force constant of the spring, \( k = 100 \text{ N/m} \)
Step 2: Substituting the values into the time period formula.
Using the formula for the time period: \[ T = 2\pi \sqrt{\frac{1}{100}} \] Simplifying the square root of the fraction: \[ \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1 \] Now, computing the final value of \( T \): \[ T = 2\pi \cdot (0.1) = 0.2\pi \text{ s} \] This directly matches option (A).
• \( T \) is the time period in seconds (\(\text{s}\))
• \( m \) is the mass in kilograms (\(\text{kg}\))
• \( k \) is the spring constant in Newtons per meter (\(\text{N/m}\))
Step 1: Identifying the given parameters from the problem statement.
From the question, we are given:
• Mass attached, \( m = 1 \text{ kg} \)
• Force constant of the spring, \( k = 100 \text{ N/m} \)
Step 2: Substituting the values into the time period formula.
Using the formula for the time period: \[ T = 2\pi \sqrt{\frac{1}{100}} \] Simplifying the square root of the fraction: \[ \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1 \] Now, computing the final value of \( T \): \[ T = 2\pi \cdot (0.1) = 0.2\pi \text{ s} \] This directly matches option (A).
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