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 \(T = 2\pi \sqrt{\frac{m}{k}}\). \[ T = 2\pi \sqrt{\frac{1}{100}} = \frac{\pi}{5} \, {s} \]