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 will use the relationship between the frequency of a vibrating string and the tension in the string. The fundamental frequency \( f \) of a wire is given by the formula:

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

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

Given:
- Length of the wire \( L = 1 \) m,
- Initial frequency \( f_1 = 256 \) Hz,
- New frequency after increasing tension \( f_2 = 320 \) Hz,
- Increase in tension by \( 1 \) kg weight, which is \( 10 \) N.

Let the initial tension be \( T_1 = T \). After increasing the tension, the new tension becomes:

\[ T_2 = T + 10 \, \text{N} \]

Using the relationship of frequencies and tensions, we can write:

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

Substituting the known values:

\[ \frac{256}{320} = \sqrt{\frac{T}{T + 10}} \]

Calculating \( \frac{256}{320} \):

\[ \frac{256}{320} = 0.8 \]

Thus, we have:

\[ 0.8 = \sqrt{\frac{T}{T + 10}} \]

Now, squaring both sides:

\[ 0.64 = \frac{T}{T + 10} \]

Cross-multiplying gives:

\[ 0.64(T + 10) = T \]

Expanding this:

\[ 0.64T + 6.4 = T \]

Rearranging the equation:

\[ T - 0.64T = 6.4 \]

This simplifies to:

\[ 0.36T = 6.4 \]

Now, solving for \( T \):

\[ T = \frac{6.4}{0.36} \approx 17.78 \, \text{N} \]

Thus, the initial tension \( T \) is approximately \( 17.78 \, \text{N} \).

### Final Answer:
The initial tension is approximately \( 17.78 \, \text{N} \).

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