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
B. `(mg)/(k)sqrt((1+2hk)/(mg))`
C. `(mg)/(k)+(mg)/(k)sqrt((1+2hk)/(mg))`
D. None of the above

Answer

To solve the problem step by step, we will analyze the situation involving a mass `m` released from a height `h` onto a spring with spring constant `k`.

### Step 1: Understand the System
The body of mass `m` is released from a height `h` and falls onto a spring. The spring will compress when the mass lands on it, and we need to find the maximum compression of the spring, which will correspond to the amplitude of vibration.

### Step 2: Energy Conservation
We will use the principle of conservation of mechanical energy. The total mechanical energy at the height `h` (potential energy) will be equal to the total mechanical energy when the spring is at maximum compression (potential energy in the spring).

1. **Potential Energy at Height h**:
\[ PE_{\text{initial}} = mgh \]

2. **Potential Energy in the Spring at Maximum Compression**:
When the spring is compressed by a distance `y`, the potential energy stored in the spring is given by:
\[ PE_{\text{spring}} = \frac{1}{2}ky^2 \]

### Step 3: Set Up the Energy Conservation Equation
At the point of maximum compression, all the potential energy from the height will be converted into the potential energy of the spring:
\[ mgh = \frac{1}{2}ky^2 \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ ky^2 = 2mgh \]
\[ y^2 = \frac{2mgh}{k} \]

### Step 5: Solve for y
Taking the square root of both sides, we find:
\[ y = \sqrt{\frac{2mgh}{k}} \]

### Step 6: Identify the Amplitude
Since `y` represents the maximum compression of the spring, it is also the amplitude of the vibration. Therefore, the amplitude `A` is given by:
\[ A = y = \sqrt{\frac{2mgh}{k}} \]

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

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