Question

Consider the following spring configurations:
If the first spring is stretched by an amount \( \Delta x \), the second system will be stretched by:

• A. \( \Delta x \)
• B. \( \frac{\Delta x}{4} \)
• C. \( \frac{\Delta x}{2} \)
• D. \( 2 \Delta x \)

Hint:

When dealing with springs in series, remember that the equivalent spring constant decreases, and the stretch is greater compared to individual springs.

Answer 1

The Correct Option is C

In this problem, we have two spring systems: the first system with spring constant \( k \) and the second system with two springs in series, each with spring constant \( k \). When two springs are arranged in series, the equivalent spring constant \( k_{\text{eq}} \) is given by: \[ \frac{1}{k_{\text{eq}}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \] Thus, the equivalent spring constant for the second system is: \[ k_{\text{eq}} = \frac{k}{2} \] Now, according to Hooke's Law: \[ F = k \Delta x \] For the first system, the force required to stretch the spring by \( \Delta x \) is: \[ F_1 = k \Delta x \] For the second system, since the equivalent spring constant is \( \frac{k}{2} \), the stretch \( \Delta x_2 \) for the same force is: \[ F_2 = k_{\text{eq}} \Delta x_2 = \frac{k}{2} \Delta x_2 \] Equating the forces in both systems, \( F_1 = F_2 \), we get: \[ k \Delta x = \frac{k}{2} \Delta x_2 \] Solving for \( \Delta x_2 \): \[ \Delta x_2 = 2 \Delta x \] Thus, the stretch for the second system is \( \frac{\Delta x}{2} \).