Question

A pan with a set of weights attached to a light spring oscillates with a time period of 0.6 s. When an additional weight is added, the time period becomes 0.7 s. The extension due to the additional weight is:

• A. 2 cm
• B. 3 cm
• C. 1 cm
• D. 4 cm

Hint:

Calculations involving the spring constant, mass, and gravitational force must be precise to correctly estimate the extensions and periods in spring-mass systems.

Answer 1

The Correct Option is B

We are given the following data: 
- The time period of the spring without the additional weight is \( T_1 = 0.6 \, {s} \), 
- The time period of the spring with the additional weight is \( T_2 = 0.7 \, {s} \). 
The time period of a mass-spring system is given by the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \( T \) is the time period, \( m \) is the mass, and \( k \) is the spring constant. 
Step 1: Find the ratio of the time periods. Using the relationship between the initial and final time periods: \[ \frac{T_2}{T_1} = \sqrt{\frac{m_2}{m_1}} \] Squaring both sides: \[ \left( \frac{T_2}{T_1} \right)^2 = \frac{m_2}{m_1} \] Substitute the values: \[ \left( \frac{0.7}{0.6} \right)^2 = \frac{m_2}{m_1} \] \[ \frac{m_2}{m_1} = \left( \frac{7}{6} \right)^2 = \frac{49}{36} \] Thus, the mass after adding the weight is \( \frac{49}{36} \) times the original mass. The additional mass is: \[ \Delta m = m_2 - m_1 = m_1 \left( \frac{49}{36} - 1 \right) = m_1 \times \frac{13}{36} \] Step 2: Find the extension. The extension of the spring \( \Delta x \) due to the additional weight is proportional to the weight \( \Delta m \). 
Using the force extension relationship: \[ F = \Delta m \cdot g = k \cdot \Delta x \] The extension \( \Delta x \) is: \[ \Delta x = \frac{\Delta m \cdot g}{k} \] Thus, the extension due to the additional weight is 3 cm. 
Conclusion: The extension due to the additional weight is 3 cm, so the correct answer is (2) 3 cm.