A body of mass m dropped from a certain height strikes a light vertical fixed spring of stifness k. the height of its fall before touching the spring the if the maximum compression of the spring the equal to `(3 mg)/k` is

Options

A. `(3mg)/(2k)`
B. `(2mg)/(k)`
C. `(3mg)/(4k)`
D. `(mg)/(4k)`

Answer

To solve the problem, we need to analyze the energy transformations that occur when a body of mass \( m \) is dropped from a height \( h \) and strikes a vertical spring with spring constant \( k \). The maximum compression of the spring is given as \( \frac{3mg}{k} \).

### Step-by-Step Solution:

1. **Identify the Energy Types**:
- When the body is dropped from height \( h \), it has gravitational potential energy which converts into kinetic energy as it falls and then into elastic potential energy when it compresses the spring.

2. **Potential Energy at Height \( h \)**:
- The gravitational potential energy (PE) of the body at height \( h \) is given by:
\[ PE = mgh \]

3. **Maximum Compression of the Spring**:
- When the body compresses the spring to its maximum compression \( x \), the potential energy stored in the spring is given by:
\[ PE_{\text{spring}} = \frac{1}{2} k x^2 \]

4. **Energy Conservation Principle**:
- At the point of maximum compression, all the gravitational potential energy is converted into the elastic potential energy of the spring. Thus, we can write:
\[ mgh = \frac{1}{2} k x^2 \]

5. **Substituting for \( x \)**:
- We know from the problem that the maximum compression \( x \) is given as \( \frac{3mg}{k} \). Substituting this into the energy conservation equation:
\[ mgh = \frac{1}{2} k \left(\frac{3mg}{k}\right)^2 \]

6. **Simplifying the Equation**:
- Substitute \( x \) into the equation:
\[ mgh = \frac{1}{2} k \cdot \frac{9m^2g^2}{k^2} \]
- This simplifies to:
\[ mgh = \frac{9mg^2}{2k} \]

7. **Solving for \( h \)**:
- Rearranging the equation to solve for \( h \):
\[ h = \frac{9g}{2k} \]

8. **Final Result**:
- Thus, the height \( h \) from which the body is dropped is:
\[ h = \frac{9mg}{2k} \]

### Conclusion:
The height \( h \) from which the body is dropped before touching the spring is \( \frac{9mg}{2k} \).