Question

A body of mass 5 kg hangs from a spring and oscillates with a time period of $2\pi$ seconds. If the ball is removed, the length of the spring will decrease by

• A. g/k metres
• B. k/g metres
• C. $ 2 \pi$ meters
• D. g metres

Answer 1

The Correct Option is D

To solve this problem, we need to understand the relationship between the oscillation of a mass-spring system and the change in length of the spring when the mass is removed.

1. Given:
   • Mass of the body, m = 5 \text{ kg}
   • Time period of oscillation, T = 2\pi \text{ seconds}
2. The formula for the time period of a mass-spring system is: T = 2\pi \sqrt{\frac{m}{k}} where k is the spring constant.
3. Rearranging the formula to solve for k gives us: k = \frac{4\pi^2 m}{T^2}
4. We can substitute the given values to find k: k = \frac{4 \pi^2 \times 5}{(2\pi)^2}
5. Simplifying this expression: k = \frac{4 \pi^2 \times 5}{4 \pi^2} = 5 \text{ N/m}
6. Now, to determine the decrease in the length of the spring when the mass is removed, consider the extension in the spring due to the weight of the mass: F = k \cdot x where x is the extension of the spring.
7. The force F is the weight of the mass, F = mg.
8. Therefore, mg = k \cdot x, and solving for x gives: x = \frac{mg}{k} = \frac{5 \times g}{5} = g \text{ metres}
9. Thus, the length of the spring will decrease by g \text{ metres} when the mass is removed, which matches the correct answer.

Hence, the correct answer is g metres.