The string of a violin has a fundamental frequency of 440 Hz. If the violin string is shortend by one fifth, itd fundamental frequency will be changed to

Options

A. 440 cps
B. 880 cps
C. 550 cps
D. 2200 cps

Answer

To solve the problem of finding the new fundamental frequency of a violin string when its length is shortened by one-fifth, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the initial conditions**:
The initial fundamental frequency \( f_1 \) is given as 440 Hz. Therefore, we have:
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = 440 \text{ Hz} \]

3. **Determine the new length**:
If the string is shortened by one-fifth, the new length \( L' \) can be calculated as:
\[ L' = L - \frac{1}{5}L = \frac{4}{5}L \]

4. **Calculate the new frequency**:
Using the new length in the frequency formula, we have:
\[ f' = \frac{1}{2L'} \sqrt{\frac{T}{\mu}} = \frac{1}{2 \times \frac{4}{5}L} \sqrt{\frac{T}{\mu}} = \frac{5}{8L} \sqrt{\frac{T}{\mu}} \]

5. **Relate the new frequency to the initial frequency**:
Since \( \frac{1}{2L} \sqrt{\frac{T}{\mu}} = f_1 \):
\[ f' = \frac{5}{4} f_1 \]
Substituting \( f_1 = 440 \text{ Hz} \):
\[ f' = \frac{5}{4} \times 440 \text{ Hz} = 550 \text{ Hz} \]

6. **Final answer**:
Therefore, the new fundamental frequency when the string is shortened by one-fifth is:
\[ f' = 550 \text{ Hz} \]