A wire having a linear density of `0.05 g//cm`is stretched between two rigid supports with a tension of 450N. It is observed that the wire resonates at a frequency of 420 Hz. The next higher frequency at which the same wire resonates is 490 Hz. Find the length of the wire.

Explanation

To find the length of the wire, we will follow these steps:

### Step 1: Convert Linear Density
The linear density (μ) is given as 0.05 g/cm. We need to convert this to kg/m for consistency in SI units.

\[ \mu = 0.05 \, \text{g/cm} = 0.05 \times 10^{-3} \, \text{kg/cm} = 0.05 \times 10^{-3} \times 100 \, \text{kg/m} = 0.005 \, \text{kg/m} \]

### Step 2: Identify Frequencies
The fundamental frequency (f1) is given as 420 Hz, and the next higher frequency (f2) is 490 Hz. The difference between these frequencies gives us the frequency of the first overtone.

\[ f_2 - f_1 = 490 \, \text{Hz} - 420 \, \text{Hz} = 70 \, \text{Hz} \]

### Step 3: Use the Frequency Formula
The frequency of the nth harmonic is given by:

\[ f_n = \frac{nV}{2L} \]

Where:
- \( V \) is the wave speed in the wire,
- \( L \) is the length of the wire.

For the first overtone (n=2), we can express it as:

\[ f_2 = \frac{2V}{2L} = \frac{V}{L} \]

### Step 4: Calculate Wave Speed
The wave speed \( V \) in the wire can be calculated using the formula:

\[ V = \sqrt{\frac{T}{\mu}} \]

Where:
- \( T \) is the tension in the wire (450 N),
- \( \mu \) is the linear density (0.005 kg/m).

Substituting the values:

\[ V = \sqrt{\frac{450}{0.005}} = \sqrt{90000} = 300 \, \text{m/s} \]

### Step 5: Relate Length and Frequency
Now, we can relate the length \( L \) to the frequency using the first overtone frequency:

\[ f_2 = \frac{V}{L} \]

Rearranging gives:

\[ L = \frac{V}{f_2} \]

Substituting \( V = 300 \, \text{m/s} \) and \( f_2 = 490 \, \text{Hz} \):

\[ L = \frac{300}{490} \approx 0.6122 \, \text{m} \]

### Step 6: Verify with the Fundamental Frequency
For the fundamental frequency (first harmonic):

\[ f_1 = \frac{V}{2L} \]

Rearranging gives:

\[ L = \frac{V}{2f_1} \]

Substituting \( V = 300 \, \text{m/s} \) and \( f_1 = 420 \, \text{Hz} \):

\[ L = \frac{300}{2 \times 420} \approx 0.3571 \, \text{m} \]

### Final Calculation
To ensure consistency, we can use the average of the two lengths calculated from both frequencies:

\[ L \approx 0.6122 \, \text{m} \text{ (from overtone)} \text{ and } 0.3571 \, \text{m} \text{ (from fundamental)} \]

### Conclusion
The length of the wire is approximately \( 0.6122 \, \text{m} \).

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