`n` waves are produced on a string in ` 1 s` . When the radius of the string is doubled and the tension is maintained the same , the number of waves produced in ` 1 s` for the same harmonic will be

Options

A. ` 2 n`
B. ` (n)/(3)`
C. `(n)/(2)`
D. `(n)/(sqrt(2))`

Explanation

To solve the problem step by step, we need to understand the relationship between the frequency of the waves produced on a string, the tension in the string, and the mass per unit length of the string.

### Step 1: Understand the relationship between frequency, tension, and mass per unit length
The frequency \( f \) of a wave on a string is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{T}{\mu}} \]
where:
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

### Step 2: Express mass per unit length in terms of radius
The mass per unit length \( \mu \) of a string can be expressed as:
\[ \mu = \frac{m}{L} = \frac{\rho \cdot A}{L} \]
where:
- \( \rho \) is the density of the material,
- \( A \) is the cross-sectional area of the string.

For a circular cross-section, the area \( A \) is given by:
\[ A = \pi r^2 \]
Thus, we can express \( \mu \) as:
\[ \mu = \frac{\rho \cdot \pi r^2}{L} \]

### Step 3: Analyze the effect of doubling the radius
When the radius \( r \) is doubled, the new radius becomes \( 2r \). The new cross-sectional area \( A' \) becomes:
\[ A' = \pi (2r)^2 = 4\pi r^2 \]
Thus, the new mass per unit length \( \mu' \) becomes:
\[ \mu' = \frac{\rho \cdot 4\pi r^2}{L} = 4\mu \]

### Step 4: Calculate the new frequency with the same tension
Since the tension \( T \) remains the same, we can substitute \( \mu' \) into the frequency formula:
\[ f' = \frac{1}{2\pi} \sqrt{\frac{T}{\mu'}} = \frac{1}{2\pi} \sqrt{\frac{T}{4\mu}} = \frac{1}{2} \cdot \frac{1}{2\pi} \sqrt{\frac{T}{\mu}} = \frac{1}{2} f \]
This shows that the new frequency \( f' \) is half of the original frequency \( f \).

### Step 5: Relate frequency to the number of waves produced
Given that \( n \) waves are produced in 1 second originally, the new number of waves produced in 1 second will be:
\[ n' = f' = \frac{1}{2} f = \frac{1}{2} n \]

### Conclusion
Therefore, when the radius of the string is doubled and the tension is maintained the same, the number of waves produced in 1 second for the same harmonic will be:
\[ \boxed{\frac{n}{2}} \]