A wire under tension vibrates with a fundamental
frequency of 600 Hz. If the length of the wire is
doubled, the radius is halved and the wire is made to
vibrate under one-ninth the tension. Then, the
fundamental frequency will became

Options

A. 400 Hz
B. 600 Hz
C. 300 Hz
D. 200 Hz

Explanation

To solve the problem, we need to analyze how the fundamental frequency of a vibrating wire changes when its length, radius, and tension are altered. The fundamental frequency \( f_0 \) of a wire is given by the formula:

\[ f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

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: Determine the initial conditions
Given:
- Initial frequency \( f_0 = 600 \, \text{Hz} \)
- Length \( L \)
- Radius \( R \)
- Tension \( T \)

### Step 2: Calculate the mass per unit length \( \mu \)
The mass per unit length \( \mu \) can be expressed as:

\[ \mu = \frac{m}{L} = \frac{\pi R^2 \rho L}{L} = \pi R^2 \rho \]

where \( \rho \) is the density of the material.

### Step 3: Analyze the changes made to the wire
1. **Length is doubled**: \( L' = 2L \)
2. **Radius is halved**: \( R' = \frac{R}{2} \)
3. **Tension is reduced to one-ninth**: \( T' = \frac{T}{9} \)

### Step 4: Calculate the new mass per unit length \( \mu' \)
The new mass per unit length \( \mu' \) becomes:

\[ \mu' = \pi (R')^2 \rho = \pi \left(\frac{R}{2}\right)^2 \rho = \pi \frac{R^2}{4} \rho \]

### Step 5: Substitute the new values into the frequency formula
Now, we can find the new fundamental frequency \( f_0' \):

\[ f_0' = \frac{1}{2L'} \sqrt{\frac{T'}{\mu'}} \]

Substituting the new values:

\[ f_0' = \frac{1}{2(2L)} \sqrt{\frac{\frac{T}{9}}{\pi \frac{R^2}{4} \rho}} \]

### Step 6: Simplify the expression
This simplifies to:

\[ f_0' = \frac{1}{4L} \sqrt{\frac{4T}{9\pi R^2 \rho}} \]

\[ f_0' = \frac{1}{4L} \cdot \frac{2}{3} \sqrt{\frac{T}{\pi R^2 \rho}} \]

\[ f_0' = \frac{1}{6} \cdot \frac{1}{2L} \sqrt{\frac{T}{\pi R^2 \rho}} \]

### Step 7: Relate it to the original frequency
Since \( \frac{1}{2L} \sqrt{\frac{T}{\pi R^2 \rho}} = f_0 \):

\[ f_0' = \frac{1}{6} f_0 \]

### Step 8: Calculate the new frequency
Substituting the original frequency:

\[ f_0' = \frac{1}{6} \cdot 600 \, \text{Hz} = 100 \, \text{Hz} \]

### Conclusion
The new fundamental frequency of the wire after the changes is:

\[ f_0' = 100 \, \text{Hz} \]