Question

The vertical extension in a light spring by a weight of 1 kg suspended from the lower end is 9.8 cm. The time period of oscillation of the spring is:

• A. \(10\pi \, {s}\)
• B. \(\frac{\pi}{5} \, {s}\)
• C. \(\frac{5\pi}{8} \, {s}\)
• D. \(\frac{\pi}{10} \, {s}\)

Hint:

For a mass-spring system, remember that the time period \(T\) is directly influenced by the mass of the object and the spring constant.

Answer 1

The Correct Option is B

Step 1: Determine the spring constant \(k\). Using Hooke's law \( F = kx \), where \(x = 0.098 \, {m}\) (extension) and \(F = mg = 1 \times 9.8 \, {N}\): \[ k = \frac{F}{x} = \frac{9.8}{0.098} = 100 \, {N/m} \] Step 2: Calculate the time period \(T\). The time period \(T\) for a mass-spring system is given by \(T = 2\pi \sqrt{\frac{m}{k}}\). \[ T = 2\pi \sqrt{\frac{1}{100}} = \frac{\pi}{5} \, {s} \]

Answer 2

To solve the problem, we need to calculate the time period of oscillation of a spring when a 1 kg weight causes an extension of 9.8 cm in the spring.

1. Understanding Hooke's Law and Simple Harmonic Motion:
The force on a spring is given by Hooke's Law:

\[ F = k \Delta x \] where: - \( F \) is the force applied to the spring, - \( k \) is the spring constant, - \( \Delta x \) is the extension of the spring. The time period \( T \) of oscillation of a mass-spring system is given by the formula: \[ T = 2 \pi \sqrt{\frac{m}{k}} \] where: - \( m \) is the mass hanging from the spring, - \( k \) is the spring constant.

2. Finding the Spring Constant:
The weight \( W \) of the 1 kg mass is given by \( W = mg \), where \( g = 9.8 \, \text{m/s}^2 \). Thus, the weight of the mass is:

\[ W = 1 \times 9.8 = 9.8 \, \text{N} \] Using Hooke's Law, the spring constant \( k \) can be found: \[ k = \frac{F}{\Delta x} = \frac{9.8 \, \text{N}}{0.098 \, \text{m}} = 100 \, \text{N/m} \]

3. Calculating the Time Period:
Now, we can substitute the values of \( m = 1 \, \text{kg} \) and \( k = 100 \, \text{N/m} \) into the time period formula:

\[ T = 2 \pi \sqrt{\frac{1}{100}} = 2 \pi \times \frac{1}{10} = \frac{\pi}{5} \, \text{s} \]

4. Conclusion:
The time period of oscillation of the spring is \( \frac{\pi}{5} \, \text{seconds} \).

Final Answer:
The correct answer is (B) \( \frac{\pi}{5} \, \text{s} \).