A sonometer wire under tension of `128 N` vibrates in resonance with a tuning fork . The vibrating portion of sonometer wire has length of `20 cm` and mass ` 1 g`. The vibrating wire at constant speed of `0.75 m//s` and an observer standing near the sonometer hear ` 1 beat//s`. Find the speed of sound in air.

Explanation

To find the speed of sound in air based on the given information, we will follow these steps:

### Step 1: Calculate the frequency of the vibrating wire (Fs)
The frequency of a vibrating string can be calculated using the formula:
\[ F_s = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the vibrating portion of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the linear mass density of the wire.

First, we need to convert the given values:
- Length \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Mass \( m = 1 \, \text{g} = 0.001 \, \text{kg} \)
- Tension \( T = 128 \, \text{N} \)

Now, calculate the linear mass density \( \mu \):
\[ \mu = \frac{m}{L} = \frac{0.001 \, \text{kg}}{0.2 \, \text{m}} = 0.005 \, \text{kg/m} \]

Now substitute the values into the frequency formula:
\[ F_s = \frac{1}{2 \times 0.2} \sqrt{\frac{128}{0.005}} = \frac{1}{0.4} \sqrt{25600} = 2.5 \times 160 = 400 \, \text{Hz} \]

### Step 2: Relate the observed frequency to the speed of sound
The observer hears a beat frequency of 1 beat per second. The apparent frequency \( F_1 \) heard by the observer when the tuning fork moves away is given by:
\[ F_1 = F_s \left( \frac{V}{V + v} \right) \]
where \( v = 0.75 \, \text{m/s} \) (the speed of the tuning fork).

Using the binomial approximation for small \( v \) compared to \( V \):
\[ F_1 \approx F_s \left( 1 - \frac{v}{V} \right) \]

The beat frequency is given by:
\[ \Delta F = F_s - F_1 = 1 \, \text{Hz} \]
Substituting for \( F_1 \):
\[ \Delta F = F_s - F_s \left( 1 - \frac{v}{V} \right) = F_s \frac{v}{V} \]
Thus, we have:
\[ 1 = F_s \frac{0.75}{V} \]

### Step 3: Solve for the speed of sound (V)
Now we can rearrange this equation to find \( V \):
\[ V = F_s \cdot 0.75 \]
Substituting \( F_s = 400 \, \text{Hz} \):
\[ V = 400 \cdot 0.75 = 300 \, \text{m/s} \]

### Final Answer
The speed of sound in air is approximately \( 300 \, \text{m/s} \).

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