A wire under tension vibrates with a frequency of 450 per second. What would be the fundamental frequency if the wire were half as long, twice as thick and under one fourth tension.

Options

A. 225 cps
B. 190 cps
C. 247 cps
D. 174 cps

Explanation

To solve the problem, we need to find the new fundamental frequency of a wire when its length, thickness, and tension are changed.

### Step-by-Step Solution:

1. **Understand the formula for fundamental frequency**:
The fundamental frequency \( f \) of a wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( \mu \) = mass per unit length of the wire

2. **Express mass per unit length**:
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{\text{density} \times \text{volume}}{L} \]
For a cylindrical wire, the volume \( V \) is given by:
\[ V = \pi r^2 L \]
Therefore,
\[ \mu = \frac{\rho \cdot \pi r^2 L}{L} = \rho \cdot \pi r^2 \]
where \( \rho \) is the density of the material.

3. **Substitute \( \mu \) back into the frequency formula**:
Substituting \( \mu \) into the frequency formula gives:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot \pi r^2}} \]

4. **Identify the new parameters**:
- The new length \( L' = \frac{L}{2} \)
- The new radius \( r' = 2r \) (twice as thick)
- The new tension \( T' = \frac{T}{4} \) (one-fourth tension)

5. **Substitute the new parameters into the frequency formula**:
The new fundamental frequency \( f' \) can be calculated as:
\[ f' = \frac{1}{2L'} \sqrt{\frac{T'}{\rho \cdot \pi (r')^2}} \]
Substituting the new values:
\[ f' = \frac{1}{2 \cdot \frac{L}{2}} \sqrt{\frac{\frac{T}{4}}{\rho \cdot \pi (2r)^2}} \]
Simplifying this:
\[ f' = \frac{1}{L} \sqrt{\frac{\frac{T}{4}}{\rho \cdot \pi \cdot 4r^2}} \]
\[ f' = \frac{1}{L} \sqrt{\frac{T}{16 \rho \cdot \pi r^2}} \]

6. **Relate \( f' \) to the original frequency \( f \)**:
We know the original frequency \( f \) is:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot \pi r^2}} \]
Therefore, we can express \( f' \) in terms of \( f \):
\[ f' = \frac{1}{L} \cdot \frac{1}{4} \cdot \sqrt{\frac{T}{\rho \cdot \pi r^2}} = \frac{f}{4} \]

7. **Calculate the new frequency**:
Given that the original frequency \( f = 450 \) Hz, we find:
\[ f' = \frac{450}{4} = 112.5 \text{ Hz} \]

### Final Answer:
The fundamental frequency of the wire when it is half as long, twice as thick, and under one-fourth tension is **112.5 Hz**.