Waves produced by two vibrators in a medium have wavelength 2 m and 2.1 m respectively. When sounded together they produce 8 beats/second. Calculate wave velocity and frequencies of the vibrators.

Answer

To solve the problem step by step, we will follow the outlined process to find the wave velocity and the frequencies of the two vibrators.

### Step 1: Understand the relationship between wave velocity, frequency, and wavelength.
The wave velocity (v) can be expressed using the formula:
\[ v = n \cdot \lambda \]
where:
- \( n \) is the frequency,
- \( \lambda \) is the wavelength.

### Step 2: Set up the equations for both vibrators.
Let:
- \( \lambda_1 = 2 \, \text{m} \) (wavelength of the first vibrator),
- \( \lambda_2 = 2.1 \, \text{m} \) (wavelength of the second vibrator).

The frequencies of the two vibrators can be expressed as:
\[ n_1 = \frac{v}{\lambda_1} \]
\[ n_2 = \frac{v}{\lambda_2} \]

### Step 3: Use the information about beats.
The problem states that the two waves produce 8 beats per second. The beat frequency is given by the absolute difference of the two frequencies:
\[ |n_1 - n_2| = 8 \]

### Step 4: Substitute the expressions for frequencies.
Substituting the expressions for \( n_1 \) and \( n_2 \):
\[ \left| \frac{v}{\lambda_1} - \frac{v}{\lambda_2} \right| = 8 \]

### Step 5: Factor out the common terms.
Factoring out \( v \):
\[ v \left| \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right| = 8 \]

### Step 6: Substitute the values of wavelengths.
Substituting \( \lambda_1 = 2 \, \text{m} \) and \( \lambda_2 = 2.1 \, \text{m} \):
\[ v \left| \frac{1}{2} - \frac{1}{2.1} \right| = 8 \]

### Step 7: Calculate the difference.
Calculating \( \frac{1}{2} - \frac{1}{2.1} \):
\[ \frac{1}{2} = 0.5 \]
\[ \frac{1}{2.1} \approx 0.4762 \]
Thus,
\[ 0.5 - 0.4762 \approx 0.0238 \]

### Step 8: Solve for wave velocity (v).
Now substituting back:
\[ v \cdot 0.0238 = 8 \]
\[ v = \frac{8}{0.0238} \approx 336 \, \text{m/s} \]

### Step 9: Calculate the frequencies of the vibrators.
Now that we have the wave velocity, we can calculate the frequencies:
\[ n_1 = \frac{v}{\lambda_1} = \frac{336}{2} = 168 \, \text{Hz} \]
\[ n_2 = \frac{v}{\lambda_2} = \frac{336}{2.1} \approx 160 \, \text{Hz} \]

### Final Results:
- Wave velocity \( v \approx 336 \, \text{m/s} \)
- Frequency of the first vibrator \( n_1 = 168 \, \text{Hz} \)
- Frequency of the second vibrator \( n_2 \approx 160 \, \text{Hz} \)