two waves of wavelengths 2 m and 2.02 m respectively, moving with the same velocity superpose to produce 2 beats per second. The velocity of the wave is

Options

A. 400.0 m/s
B. 404.4 m/s
C. 402.2 m/s
D. 406.0 m/s

Explanation

To solve the problem, we need to find the velocity of the waves given their wavelengths and the number of beats produced. Here’s a step-by-step solution:

### Step 1: Understand the relationship between frequency, wavelength, and velocity
The frequency \( f \) of a wave is related to its velocity \( v \) and wavelength \( \lambda \) by the formula:
\[ f = \frac{v}{\lambda} \]

### Step 2: Define the frequencies of the two waves
Let:
- \( \lambda_1 = 2 \, \text{m} \) (wavelength of the first wave)
- \( \lambda_2 = 2.02 \, \text{m} \) (wavelength of the second wave)

Using the relationship from Step 1, we can express the frequencies:
\[ f_1 = \frac{v}{\lambda_1} = \frac{v}{2} \]
\[ f_2 = \frac{v}{\lambda_2} = \frac{v}{2.02} \]

### Step 3: Use the beat frequency formula
The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f_1 - f_2| = 2 \, \text{beats per second} \]
Thus, we can write:
\[ \left| \frac{v}{2} - \frac{v}{2.02} \right| = 2 \]

### Step 4: Simplify the equation
To simplify, we can express the left side as:
\[ \frac{v}{2} - \frac{v}{2.02} = 2 \]
Finding a common denominator (which is \( 2 \times 2.02 \)):
\[ \frac{v \cdot 2.02 - v \cdot 2}{2 \cdot 2.02} = 2 \]
This simplifies to:
\[ \frac{v(2.02 - 2)}{2 \cdot 2.02} = 2 \]
\[ \frac{v(0.02)}{4.04} = 2 \]

### Step 5: Solve for \( v \)
Now, we can solve for \( v \):
\[ v(0.02) = 2 \cdot 4.04 \]
\[ v(0.02) = 8.08 \]
\[ v = \frac{8.08}{0.02} \]
\[ v = 404 \, \text{m/s} \]

### Final Answer
The velocity of the wave is:
\[ \boxed{404 \, \text{m/s}} \]