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

Options

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

Explanation

To solve the problem, we need to understand how the fundamental frequency of a vibrating wire is affected by changes in its length, thickness (diameter), and tension. The fundamental frequency \( f \) of a wire is given by the formula:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the mass per unit length of the wire.

### Step 1: Identify the initial conditions
- Initial frequency \( f_i = 450 \, \text{Hz} \)
- Let the initial length \( L_i \), initial tension \( T_i \), and initial diameter \( D_i \) be the parameters we will modify.

### Step 2: Determine the new conditions
- New length \( L_f = \frac{L_i}{2} \) (half as long)
- New diameter \( D_f = 2D_i \) (twice as thick)
- New tension \( T_f = \frac{T_i}{4} \) (one fourth tension)

### Step 3: Calculate the mass per unit length
The mass per unit length \( \mu \) is given by:

\[ \mu = \frac{m}{L} = \rho \cdot A \]

where \( A \) is the cross-sectional area of the wire. The area \( A \) for a circular wire is:

\[ A = \frac{\pi D^2}{4} \]

Thus, the new mass per unit length \( \mu_f \) can be expressed as:

\[ \mu_f = \rho \cdot A_f = \rho \cdot \frac{\pi (D_f)^2}{4} = \rho \cdot \frac{\pi (2D_i)^2}{4} = \rho \cdot \frac{\pi \cdot 4D_i^2}{4} = \rho \cdot \pi D_i^2 = 4 \mu_i \]

### Step 4: Substitute the new values into the frequency formula
Now we can substitute the new values into the frequency formula:

\[ f_f = \frac{1}{2L_f} \sqrt{\frac{T_f}{\mu_f}} = \frac{1}{2 \cdot \frac{L_i}{2}} \sqrt{\frac{\frac{T_i}{4}}{4\mu_i}} = \frac{1}{L_i} \sqrt{\frac{T_i}{16\mu_i}} = \frac{1}{L_i} \cdot \frac{1}{4} \sqrt{\frac{T_i}{\mu_i}} \]

### Step 5: Relate the new frequency to the initial frequency
From the initial frequency formula, we have:

\[ f_i = \frac{1}{2L_i} \sqrt{\frac{T_i}{\mu_i}} \]

Thus, we can express the new frequency in terms of the initial frequency:

\[ f_f = \frac{1}{4} f_i \]

### Step 6: Calculate the final frequency
Now substituting the initial frequency:

\[ f_f = \frac{1}{4} \cdot 450 \, \text{Hz} = 112.5 \, \text{Hz} \]

### Final Answer
The fundamental frequency of the wire under the new conditions would be **112.5 Hz**.