A wire having a linear density `0.1 kg//m` is kept under a tension of 490N. It is observed that it resonates at a frequency of 400 Hz. The next higher frequency is 450 Hz. Find the length of the wire.

Options

A. (a)0.4 m
B. (b)0.7m
C. (c)0.6m
D. (d)0.49m

Explanation

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

### Step 1: Identify the fundamental frequency
The problem states that the wire resonates at two frequencies: 400 Hz and 450 Hz. The difference between these two frequencies gives us the fundamental frequency (first harmonic).

\[ f_1 = 450 \, \text{Hz} - 400 \, \text{Hz} = 50 \, \text{Hz} \]

### Step 2: Use the formula for the fundamental frequency
The formula for the fundamental frequency \( f \) of a vibrating string is given by:

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

where \( V \) is the wave speed and \( L \) is the length of the wire.

### Step 3: Calculate the wave speed \( V \)
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 and \( \mu \) is the linear density.

Given:
- Tension \( T = 490 \, \text{N} \)
- Linear density \( \mu = 0.1 \, \text{kg/m} \)

Substituting these values into the formula for \( V \):

\[ V = \sqrt{\frac{490 \, \text{N}}{0.1 \, \text{kg/m}}} = \sqrt{4900 \, \text{m}^2/\text{s}^2} = 70 \, \text{m/s} \]

### Step 4: Substitute \( V \) into the fundamental frequency formula
Now we can substitute \( V \) back into the fundamental frequency formula to find the length \( L \):

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

Rearranging for \( L \):

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

Substituting the values we have:

\[ L = \frac{70 \, \text{m/s}}{2 \times 50 \, \text{Hz}} = \frac{70}{100} = 0.7 \, \text{m} \]

### Final Answer
The length of the wire is:

\[ L = 0.7 \, \text{m} \]

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