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 from an unstretched position, it gains kinetic energy as it falls, which is then converted into elastic potential energy. The "maximum extension" occurs at the point where the instantaneous velocity of the mass becomes zero.

Step 2: Key Formula or Approach:
We use the Law of Conservation of Mechanical Energy. Let \(x\) be the maximum extension. Loss in Gravitational Potential Energy = Gain in Elastic Potential Energy.

Step 3: Detailed Explanation:
1. The mass descends by a distance \(x\), so the loss in potential energy is \(Mgx\). 2. The spring is stretched by \(x\), so the gain in elastic potential energy is \(\frac{1}{2}kx^2\). Equating the two: \[ Mgx = \frac{1}{2}kx^2 \] Since \(x \neq 0\), we can divide both sides by \(x\): \[ Mg = \frac{1}{2}kx \] \[ x = \frac{2Mg}{k} \]

Step 4: Final Answer
The maximum extension in the spring is \(2Mg/k\).