A piano wire having a diameter of `0.90 mm` is replaced by another wire of the same material but with a diameter of `0.93 mm`.If the tension of the wire is kept the same , then the percentage change in the frequency of the fundamental tone is

Options

A. ` + 3 %`
B. `+ 3.2 %`
C. `- 3.2 %`
D. `- 3%`

Explanation

To solve the problem of finding the percentage change in the frequency of the fundamental tone when a piano wire is replaced with another wire of a slightly larger diameter, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between frequency and linear mass density (μ)**:
The frequency (f) of the fundamental tone 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
- \( \mu \) = linear mass density of the wire

2. **Express linear mass density (μ) in terms of diameter (D)**:
The linear mass density can be expressed as:
\[ \mu = \frac{m}{L} = \frac{\text{Density} \times \text{Volume}}{L} \]
The volume of the wire can be expressed in terms of its cross-sectional area:
\[ \text{Volume} = \text{Area} \times \text{Length} = \frac{\pi D^2}{4} \times L \]
Thus, the linear mass density becomes:
\[ \mu = \frac{\rho \cdot \frac{\pi D^2}{4} \cdot L}{L} = \frac{\rho \pi D^2}{4} \]
where \( \rho \) is the density of the material.

3. **Substitute μ into the frequency formula**:
Replacing μ in the frequency formula gives:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\frac{\rho \pi D^2}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{\rho \pi D^2}} = \frac{2}{L} \sqrt{\frac{T}{\rho \pi}} \cdot \frac{1}{D} \]
This shows that frequency is inversely proportional to the diameter (D) of the wire.

4. **Calculate the change in diameter**:
The initial diameter \( D_1 = 0.90 \, \text{mm} \) and the new diameter \( D_2 = 0.93 \, \text{mm} \).
The change in diameter \( \Delta D = D_2 - D_1 = 0.93 \, \text{mm} - 0.90 \, \text{mm} = 0.03 \, \text{mm} \).

5. **Calculate the percentage change in diameter**:
The percentage change in diameter is given by:
\[ \text{Percentage Change} = \frac{\Delta D}{D_1} \times 100 = \frac{0.03 \, \text{mm}}{0.90 \, \text{mm}} \times 100 \]
\[ = \frac{3}{90} \times 100 = 3.33\% \]

6. **Determine the percentage change in frequency**:
Since frequency is inversely proportional to diameter, the percentage change in frequency can be calculated as:
\[ \text{Percentage Change in Frequency} = -\frac{\Delta D}{D_1} \times 100 = -3.33\% \]

### Final Answer:
The percentage change in the frequency of the fundamental tone is approximately **-3.33%**.