A body of mass `m` is released from a height h to a scale pan hung from a spring. The spring constant of the spring is `k`, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration is

Options

A. `mg/k`
B. `mg/ksqrt((1+(2hk)/mg))`
C. `mg/k+mg/ksqrt((1+(2hk)/mg))`
D. `mg/ksqrt((1+2hk)/(mg) (mg)/k))`

Explanation

To solve the problem, we will use the principle of conservation of energy. The potential energy lost by the mass when it falls will be equal to the elastic potential energy stored in the spring when it is compressed.

### Step-by-Step Solution:

1. **Identify the Potential Energy Lost**:
When the body of mass \( m \) is released from height \( h \), the potential energy lost when it falls to the spring is given by:
\[ \Delta PE = mgh \]

2. **Identify the Elastic Potential Energy Gained**:
When the mass compresses the spring by a distance \( y \), the elastic potential energy stored in the spring is:
\[ PE_{spring} = \frac{1}{2} k y^2 \]

3. **Apply Conservation of Energy**:
According to the conservation of energy, the potential energy lost by the mass will be equal to the elastic potential energy gained by the spring:
\[ mgh = \frac{1}{2} k y^2 \]

4. **Rearranging the Equation**:
Rearranging the equation to solve for \( y \):
\[ k y^2 = 2mgh \]
\[ y^2 = \frac{2mgh}{k} \]
\[ y = \sqrt{\frac{2mgh}{k}} \]

5. **Identify the Amplitude of Vibration**:
The amplitude of vibration \( A \) is equal to the maximum displacement \( y \) when the mass compresses the spring. Therefore, the amplitude of vibration is:
\[ A = y = \sqrt{\frac{2mgh}{k}} \]

### Final Answer:
The amplitude of vibration is:
\[ A = \sqrt{\frac{2mgh}{k}} \]