A string of mass per unit length `mu = 6 × 10^(-3)` kg/m is fixed at both ends under the tension 540 N. If the string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the string?

Options

A. `8.1 m`
B. `5.1m`
C. `2.1m`
D. `1.1m`

Answer

To find the length of the string that is fixed at both ends and is vibrating in resonance at two consecutive frequencies, we can follow these steps:

### Step 1: Understand the relationship between frequency, tension, and length of the string.
The frequency of a vibrating string fixed at both ends is given by the formula:
\[ f_n = \frac{nV}{2L} \]
where:
- \( f_n \) is the frequency,
- \( n \) is the harmonic number (1 for fundamental frequency, 2 for first overtone, etc.),
- \( V \) is the wave velocity,
- \( L \) is the length of the string.

### Step 2: Determine the wave velocity \( V \).
The wave velocity \( V \) on the string can be calculated using the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
where:
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

Given:
- \( T = 540 \, \text{N} \)
- \( \mu = 6 \times 10^{-3} \, \text{kg/m} \)

Substituting the values:
\[ V = \sqrt{\frac{540}{6 \times 10^{-3}}} \]

### Step 3: Calculate \( V \).
Calculating \( V \):
\[ V = \sqrt{\frac{540}{0.006}} = \sqrt{90000} = 300 \, \text{m/s} \]

### Step 4: Use the frequencies to find the length \( L \).
Let the two consecutive frequencies be \( f_1 = 420 \, \text{Hz} \) and \( f_2 = 490 \, \text{Hz} \). The difference between these frequencies is:
\[ f_2 - f_1 = 490 - 420 = 70 \, \text{Hz} \]

From the frequency formula, we have:
\[ f_2 - f_1 = \frac{V}{2L} \]
Substituting the known values:
\[ 70 = \frac{300}{2L} \]

### Step 5: Solve for \( L \).
Rearranging the equation to find \( L \):
\[ 2L = \frac{300}{70} \]
\[ L = \frac{300}{140} = \frac{30}{14} \approx 2.14 \, \text{m} \]

### Conclusion
Thus, the length of the string is approximately:
\[ L \approx 2.14 \, \text{m} \]