A wire of length one metre under a certain initial tension emits a sound of fundamental frequency 256 Hz. When the tesnion is increased by 1 kg wt , the frequency of the fundamental node increases to 320 Hz. The initial tension is

Options

A. `3//4 kg ` wt
B. 4/3 kg wt
C. 16/9 kg wt
D. 20/9 kg wt

Answer

To solve the problem, we need to find the initial tension \( T_1 \) in the wire. We are given the following information:

- Length of the wire \( L = 1 \, \text{m} \)
- Initial frequency \( f_1 = 256 \, \text{Hz} \)
- Frequency after increasing the tension \( f_2 = 320 \, \text{Hz} \)
- Increase in tension \( \Delta T = 1 \, \text{kg wt} \)

### Step 1: Understand the relationship between frequency and tension

The fundamental frequency of a vibrating string is given by the formula:

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

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

Since the length \( L \) and the linear mass density \( \mu \) are constant, we can say that the frequency is directly proportional to the square root of the tension:

\[ f \propto \sqrt{T} \]

### Step 2: Set up the ratio of frequencies

From the relationship established, we can write:

\[ \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} \]

Where:
- \( T_1 = T \) (initial tension)
- \( T_2 = T + 1 \) (tension after increasing by 1 kg)

Substituting the known frequencies:

\[ \frac{320}{256} = \sqrt{\frac{T + 1}{T}} \]

### Step 3: Simplify the ratio

Calculating the left side:

\[ \frac{320}{256} = \frac{5}{4} \]

Thus, we have:

\[ \frac{5}{4} = \sqrt{\frac{T + 1}{T}} \]

### Step 4: Square both sides

Squaring both sides gives:

\[ \left(\frac{5}{4}\right)^2 = \frac{T + 1}{T} \]

Calculating the square:

\[ \frac{25}{16} = \frac{T + 1}{T} \]

### Step 5: Cross-multiply to solve for \( T \)

Cross-multiplying gives:

\[ 25T = 16(T + 1) \]

Expanding the right side:

\[ 25T = 16T + 16 \]

### Step 6: Rearrange the equation

Rearranging gives:

\[ 25T - 16T = 16 \]

This simplifies to:

\[ 9T = 16 \]

### Step 7: Solve for \( T \)

Dividing both sides by 9 gives:

\[ T = \frac{16}{9} \, \text{kg wt} \]

### Conclusion

The initial tension \( T_1 \) in the wire is \( \frac{16}{9} \, \text{kg wt} \).

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