Question

An ideal spring with spring-constant \(k\) is hung from the ceiling and a mass \(M\) is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is

• A. \(4Mg/k\)
• B. \(2Mg/k\)
• C. \(Mg/k\)
• D. \(Mg/2k\)

Hint:

Don't confuse "equilibrium position" with "maximum extension." At equilibrium, \(F = kx = Mg\), so \(x = Mg/k\). But due to momentum, the mass overshoots this point and reaches exactly double that distance!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept: 
When the mass is released, it falls under gravity, and the spring stretches. The mass will momentarily come to rest at the maximum extension $x_{\max}$. At this point, the loss in gravitational potential energy is completely converted into the elastic potential energy of the spring. 
Step 2: Key Formula or Approach: 
Using the Work-Energy Theorem or Conservation of Mechanical Energy: \[ \Delta K + \Delta U_g + \Delta U_s = 0 \] Since the initial and final velocities are zero, $\Delta K = 0$. 
Step 3: Detailed Explanation: 
Let the maximum extension be $x_{\max}$. The loss in gravitational potential energy of the mass is: \[ \Delta U_g = -Mg x_{\max} \] The gain in elastic potential energy of the spring is: \[ \Delta U_s = \frac{1}{2} k x_{\max}^2 \] Equating the magnitudes (or setting the sum to zero): \[ Mg x_{\max} = \frac{1}{2} k x_{\max}^2 \] Since $x_{\max} \neq 0$: \[ Mg = \frac{1}{2} k x_{\max} \] \[ x_{\max} = \frac{2Mg}{k} \] Step 4: Final Answer: 
The maximum extension in the spring is $2Mg/k$.