Question

When the mass attached to a spring is increased from 4 kg to 9 kg, the time period of oscillation increases by \(0.2\pi\) s. Then the spring constant of the spring is

• A. \(80~\text{N m}^{-1}\)
• B. \(200~\text{N m}^{-1}\)
• C. \(50~\text{N m}^{-1}\)
• D. \(100~\text{N m}^{-1}\)

Hint:

Use the time period formula \(T = 2\pi \sqrt{m/k}\) for two different masses and solve for \(k\) using the given change in time period.

Answer 1

The Correct Option is D

Step 1: Time period of a spring-mass system is
\[ T = 2\pi \sqrt{\frac{m}{k}} \] Step 2: Apply time period formula for both masses
Let \(T_1\) and \(T_2\) be the time periods for masses 4 kg and 9 kg, respectively. Then,
\[ T_1 = 2\pi \sqrt{\frac{4}{k}}, \quad T_2 = 2\pi \sqrt{\frac{9}{k}} \] Given: \(T_2 - T_1 = 0.2\pi\)
\[ 2\pi\left(\sqrt{\frac{9}{k}} - \sqrt{\frac{4}{k}}\right) = 0.2\pi \Rightarrow 2\left(\sqrt{9} - \sqrt{4}\right)\cdot \frac{1}{\sqrt{k}} = 0.2 \Rightarrow 2(3 - 2)\cdot \frac{1}{\sqrt{k}} = 0.2 \] \[ \frac{2}{\sqrt{k}} = 0.2 \Rightarrow \sqrt{k} = 10 \Rightarrow k = 100~\text{N m}^{-1} \]