A wire having a linear mass density `5.0xx10^(3) kg//m` is stretched between two rigid supports with tension of 450 N. The wire resonates at a frequency of `420 Hz`. The next higher frequency at which the same wire resonates is` 480 N`. The length of the wire is

Options

A. `2.0m`
B. `2.1m`
C. `2.5m`
D. `3m`

Explanation

To find the length of the wire, we can use the formula for the frequency of a vibrating string fixed at both ends:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where:
- \( f \) is the frequency,
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the linear mass density of the wire.

### Step 1: Identify the given values
- Linear mass density (\( \mu \)) = \( 5.0 \times 10^{-3} \, \text{kg/m} \)
- Tension (\( T \)) = \( 450 \, \text{N} \)
- First frequency (\( f_1 \)) = \( 420 \, \text{Hz} \)
- Second frequency (\( f_2 \)) = \( 480 \, \text{Hz} \)

### Step 2: Calculate the fundamental frequency
The difference between the two frequencies gives us the fundamental frequency:

\[ f_0 = f_2 - f_1 = 480 \, \text{Hz} - 420 \, \text{Hz} = 60 \, \text{Hz} \]

### Step 3: Use the frequency formula to find the length
We can rearrange the frequency formula to solve for \( L \):

\[ L = \frac{1}{2f} \sqrt{\frac{T}{\mu}} \]

Substituting the fundamental frequency \( f_0 = 60 \, \text{Hz} \):

\[ L = \frac{1}{2 \times 60} \sqrt{\frac{450}{5.0 \times 10^{-3}}} \]

### Step 4: Calculate the square root term
First, calculate \( \frac{450}{5.0 \times 10^{-3}} \):

\[ \frac{450}{5.0 \times 10^{-3}} = 450 \div 0.005 = 90000 \]

Now, take the square root:

\[ \sqrt{90000} = 300 \]

### Step 5: Substitute back to find \( L \)
Now substitute back into the equation for \( L \):

\[ L = \frac{1}{120} \times 300 = \frac{300}{120} = 2.5 \, \text{m} \]

### Final Answer
The length of the wire is \( 2.5 \, \text{m} \).