Wavelengths of two notes in air are 80/175 m and 80/173 m. Each note produces 4 beats/s, with a third note of a fixed frequency. The speed of sound in air is

Options

A. 400 m/s
B. 300 m/s
C. 280 m/s
D. 320 m/s

Answer

To solve the problem, we need to find the speed of sound in air given the wavelengths of two notes and the number of beats produced per second. Here’s the step-by-step solution:

### Step 1: Identify the given data
We have two wavelengths:
- \( \lambda_1 = \frac{80}{175} \, \text{m} \)
- \( \lambda_2 = \frac{80}{173} \, \text{m} \)

The number of beats produced per second is given as:
- Beats per second = 4

### Step 2: Relate frequency and wavelength
The speed of sound \( v \) can be expressed in terms of frequency \( f \) and wavelength \( \lambda \):
\[ v = f \cdot \lambda \]

For the two notes, we can write:
- \( v = f_1 \cdot \lambda_1 \)
- \( v = f_2 \cdot \lambda_2 \)

Where \( f_1 \) and \( f_2 \) are the frequencies corresponding to \( \lambda_1 \) and \( \lambda_2 \) respectively.

### Step 3: Express frequencies in terms of speed and wavelength
From the above equations, we can express the frequencies as:
- \( f_1 = \frac{v}{\lambda_1} \)
- \( f_2 = \frac{v}{\lambda_2} \)

### Step 4: Determine the relationship between the frequencies
Since the two notes produce 4 beats per second with a third note of fixed frequency \( f \), we have:
- \( |f_1 - f| = 4 \)
- \( |f - f_2| = 4 \)

This implies:
- \( f_1 - f = 4 \) and \( f - f_2 = 4 \) (assuming \( f_1 > f > f_2 \))

Adding these two equations gives:
\[ f_1 - f_2 = 8 \]

### Step 5: Substitute frequencies in terms of speed and wavelength
Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{v}{\lambda_1} - \frac{v}{\lambda_2} = 8 \]

### Step 6: Factor out \( v \)
Factoring out \( v \) gives:
\[ v \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = 8 \]

### Step 7: Substitute the values of wavelengths
Substituting the values of \( \lambda_1 \) and \( \lambda_2 \):
\[ \lambda_1 = \frac{80}{175} \quad \text{and} \quad \lambda_2 = \frac{80}{173} \]

Calculating \( \frac{1}{\lambda_1} \) and \( \frac{1}{\lambda_2} \):
\[ \frac{1}{\lambda_1} = \frac{175}{80} \quad \text{and} \quad \frac{1}{\lambda_2} = \frac{173}{80} \]

Thus:
\[ \frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{175}{80} - \frac{173}{80} = \frac{2}{80} = \frac{1}{40} \]

### Step 8: Substitute back to find \( v \)
Now substituting back into the equation:
\[ v \left( \frac{1}{40} \right) = 8 \]
\[ v = 8 \times 40 = 320 \, \text{m/s} \]

### Conclusion
The speed of sound in air is:
\[ \boxed{320 \, \text{m/s}} \]

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