A uniform string is vibrating with a fundamental frequency 'f'. The new frequency, if radius & length both are doubled would be

Options

A. 2f
B. 3f
C. `f/4`
D. `f/3`

Explanation

To solve the problem step by step, we need to analyze how the frequency of a vibrating string changes when both its length and radius are doubled.

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

### Step 2: Determine the mass per unit length \( \mu \)
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \frac{m}{L} = \text{density} \times \text{cross-sectional area} \]
For a cylindrical string, the cross-sectional area \( A \) is given by:
\[ A = \pi r^2 \]
Thus, we can write:
\[ \mu = \text{density} \times \pi r^2 \]

### Step 3: Substitute \( \mu \) into the frequency formula
Substituting \( \mu \) into the frequency formula gives:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\text{density} \times \pi r^2}} \]

### Step 4: Analyze the effect of doubling the length and radius
If the length \( L \) and radius \( r \) are both doubled, we have:
- New length \( L' = 2L \)
- New radius \( r' = 2r \)

Now, we can find the new mass per unit length \( \mu' \):
\[ \mu' = \text{density} \times \pi (r')^2 = \text{density} \times \pi (2r)^2 = \text{density} \times \pi \times 4r^2 = 4 \mu \]

### Step 5: Calculate the new frequency \( f' \)
Substituting the new values into the frequency formula:
\[ f' = \frac{1}{2L'} \sqrt{\frac{T}{\mu'}} = \frac{1}{2(2L)} \sqrt{\frac{T}{4\mu}} = \frac{1}{4L} \sqrt{\frac{T}{\mu}} \]
This shows that:
\[ f' = \frac{1}{4} f \]

### Conclusion
The new frequency when both the radius and length of the string are doubled is:
\[ f' = \frac{f}{4} \]