Question

A load of mass \(m\) falls from a height \(h\) onto the scale pan hung from a spring of mass \(m\) and force constant \(k\). If the spring constant is such that the scale pan is zero and the mass does not bounce relative to the pan, then the amplitude of vibration is: 

• A. \(\dfrac{mg}{k}\)
• B. \(\dfrac{mg}{k}\sqrt{\dfrac{1+2hk}{mg}}\)
• C. \(\dfrac{mg}{k}+\dfrac{mg}{k}\sqrt{\dfrac{1+2hk}{mg}}\)
• D. \(\dfrac{mg}{k}\sqrt{\dfrac{1+2hk}{mg}-\dfrac{k}{}}\)

Hint:

For spring–mass impact problems: \[ \text{Amplitude} = x_{\max} - x_{\text{equilibrium}} \] Always shift reference to equilibrium position.

Answer 1

The Correct Option is B

Step 1: Initial kinetic energy just before impact: \[ K = mgh \]
Step 2: Maximum extension \(x\) satisfies energy conservation: \[ mgh = \frac{1}{2}kx^2 - mgx \]
Step 3: Solve quadratic to find maximum extension.
Step 4: Amplitude of vibration is measured from equilibrium position: \[ A = x - \frac{mg}{k} \]
Step 5: Substituting gives: \[ A=\frac{mg}{k}\sqrt{\frac{1+2hk}{mg}} \]