The frequency of a stetched unifrom wire under tension is in resonance with the fundamental frequency of closed tube. If the tension in the wire is increased by 8 N, it is in resonance with the first overtone of the closed tube. The initial tension in the wire is

Options

A. 1N
B. 4N
C. 8N
D. 16N

Answer

To solve the problem, we need to find the initial tension in the wire, denoted as \( T \). The problem states that the frequency of the stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. When the tension is increased by 8 N, the wire resonates with the first overtone of the closed tube.

Let's break down the solution step by step:

### Step 1: Define the Initial Conditions
Let the initial tension in the wire be \( T \). The frequency of the wire under this tension can be expressed as:
\[ F_W \propto \sqrt{T} \]
This frequency is in resonance with the fundamental frequency of the closed tube, which we can denote as:
\[ F_T = \frac{V}{4L} \]
where \( V \) is the velocity of sound and \( L \) is the length of the tube.

### Step 2: Define the New Conditions After Increasing Tension
When the tension in the wire is increased by 8 N, the new tension becomes:
\[ T' = T + 8 \]
The frequency of the wire under this new tension is:
\[ F'_W \propto \sqrt{T + 8} \]
This frequency is now in resonance with the first overtone of the closed tube, which can be expressed as:
\[ F'_T = \frac{3V}{4L} \]

### Step 3: Set Up the Resonance Conditions
Since both frequencies are in resonance, we can set up the following equations:
1. \( F_W \propto \sqrt{T} \)
2. \( F_T = \frac{V}{4L} \)
3. \( F'_W \propto \sqrt{T + 8} \)
4. \( F'_T = \frac{3V}{4L} \)

From the resonance conditions, we have:
\[ \frac{V}{4L} \propto \sqrt{T} \]
\[ \frac{3V}{4L} \propto \sqrt{T + 8} \]

### Step 4: Formulate the Ratios
Taking the ratio of these two equations gives us:
\[ \frac{V/4L}{3V/4L} = \frac{\sqrt{T}}{\sqrt{T + 8}} \]
This simplifies to:
\[ \frac{1}{3} = \frac{\sqrt{T}}{\sqrt{T + 8}} \]

### Step 5: Square Both Sides
Squaring both sides results in:
\[ \left(\frac{1}{3}\right)^2 = \frac{T}{T + 8} \]
\[ \frac{1}{9} = \frac{T}{T + 8} \]

### Step 6: Cross-Multiply and Solve for T
Cross-multiplying gives:
\[ 1 \cdot (T + 8) = 9T \]
\[ T + 8 = 9T \]
Rearranging this leads to:
\[ 8T = 8 \]
Thus, we find:
\[ T = 1 \text{ N} \]

### Final Answer
The initial tension in the wire is \( T = 1 \text{ N} \).

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