Two sound waves with wavelengths `5.0 m` and `5.5 m` respectively, each propagates in a gas with velocity `330m//s` We expect the following number of beats per second:

Options

A. 12
B. zero
C. 1
D. 6

Answer

To solve the problem of finding the number of beats per second produced by two sound waves with different wavelengths, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Wavelength of the first wave, \( \lambda_1 = 5.0 \, \text{m} \)
- Wavelength of the second wave, \( \lambda_2 = 5.5 \, \text{m} \)
- Velocity of sound in gas, \( v = 330 \, \text{m/s} \)

2. **Calculate the Frequency of the First Wave:**
- The formula for frequency (\( f \)) is given by:
\[ f = \frac{v}{\lambda} \]
- For the first wave:
\[ f_1 = \frac{330 \, \text{m/s}}{5.0 \, \text{m}} = 66 \, \text{Hz} \]

3. **Calculate the Frequency of the Second Wave:**
- Using the same formula for the second wave:
\[ f_2 = \frac{330 \, \text{m/s}}{5.5 \, \text{m}} = 60 \, \text{Hz} \]

4. **Calculate the Number of Beats per Second:**
- The number of beats per second is given by the absolute difference in frequencies:
\[ \text{Number of beats} = |f_1 - f_2| = |66 \, \text{Hz} - 60 \, \text{Hz}| = 6 \, \text{Hz} \]

5. **Conclusion:**
- The expected number of beats per second is \( 6 \, \text{Hz} \).

### Final Answer:
The expected number of beats per second is **6**.

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