A point mass m is suspended at the end of a massless wire of length l and cross section. If Y is the Young's modulus for the wire, obtain the frequency of oscillation for the simple harmonic motion along the vertical line.

Explanation

To find the frequency of oscillation for a point mass \( m \) suspended at the end of a massless wire of length \( l \) and cross-section area \( A \) with Young's modulus \( Y \), we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Setup**:
- A point mass \( m \) is suspended from a wire of length \( L \) and cross-sectional area \( A \). The wire is under tension due to the weight of the mass, which is \( mg \) (where \( g \) is the acceleration due to gravity).

2. **Initial Extension of the Wire**:
- The Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Here, \( F \) is the force (tension \( T \)), \( \Delta L \) is the extension of the wire, and \( L \) is the original length of the wire.
- When the mass is at rest, the tension in the wire is equal to the weight of the mass:
\[ T = mg \]
- Therefore, we can write:
\[ Y = \frac{mg/A}{\Delta L/L} \]
- Rearranging gives us:
\[ \Delta L = \frac{mgL}{YA} \]

3. **Considering Oscillations**:
- When the mass is displaced slightly from its equilibrium position, it will experience a restoring force due to the tension in the wire.
- The new length of the wire when the mass is displaced by a small distance \( x \) will be \( L + \Delta L + x \).

4. **Restoring Force**:
- The new tension in the wire when displaced is:
\[ T' = Y \frac{A}{L} \left( \Delta L + x \right) \]
- The restoring force \( F \) acting on the mass is given by:
\[ F = T' - mg = Y \frac{A}{L} \left( \Delta L + x \right) - mg \]
- Substituting \( \Delta L \):
\[ F = Y \frac{A}{L} \left( \frac{mgL}{YA} + x \right) - mg \]
- Simplifying, we find:
\[ F = Y \frac{A}{L} \frac{mgL}{YA} + Y \frac{A}{L} x - mg \]
- The first term cancels out with \( mg \), leaving:
\[ F = Y \frac{A}{L} x \]

5. **Relating to Simple Harmonic Motion**:
- For simple harmonic motion, the restoring force is proportional to the displacement:
\[ F = -kx \]
- Here, \( k = Y \frac{A}{L} \).

6. **Finding the Angular Frequency**:
- The equation of motion for SHM gives us:
\[ m \frac{d^2x}{dt^2} = -kx \]
- This leads to:
\[ \omega^2 = \frac{k}{m} = \frac{Y A}{L m} \]
- Therefore, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{Y A}{L m}} \]

7. **Calculating the Frequency**:
- The frequency \( f \) is related to the angular frequency by:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{Y A}{L m}} \]

### Final Answer
The frequency of oscillation for the simple harmonic motion along the vertical line is:
\[ f = \frac{1}{2\pi} \sqrt{\frac{Y A}{L m}} \]