Two plane harmonic sound waves are expressed by the equations.
`y_(1)(x,t)-A cos (0.5 pi x-100 pit), y_(2)(x,t)=A cos(0.46 pix-92pi t)` (All parameters are in MKS) :
How many times does an observer hear maximum intensity in one second :-

Options

A. `4`
B. `6`
C. `8`
D. `10`

Explanation

To solve the problem, we need to determine how many times an observer hears maximum intensity in one second when two harmonic sound waves interfere. This phenomenon is known as "beats," which occur due to the superposition of two waves with slightly different frequencies.

### Step-by-step Solution:

1. **Identify the wave equations**:
The two given wave equations are:
\[ y_1(x,t) = A \cos(0.5 \pi x - 100 \pi t) \]
\[ y_2(x,t) = A \cos(0.46 \pi x - 92 \pi t) \]

2. **Determine the angular frequencies (\(\omega\))**:
The angular frequency \(\omega\) is the coefficient of \(t\) in the wave equation.
- For the first wave:
\[ \omega_1 = 100 \pi \]
- For the second wave:
\[ \omega_2 = 92 \pi \]

3. **Calculate the frequencies (\(f\))**:
The relationship between angular frequency and frequency is given by:
\[ \omega = 2 \pi f \]
Thus, we can find the frequencies:
- For the first wave:
\[ 100 \pi = 2 \pi f_1 \implies f_1 = \frac{100 \pi}{2 \pi} = 50 \text{ Hz} \]
- For the second wave:
\[ 92 \pi = 2 \pi f_2 \implies f_2 = \frac{92 \pi}{2 \pi} = 46 \text{ Hz} \]

4. **Calculate the beat frequency**:
The beat frequency \(N\) is given by the absolute difference in frequencies:
\[ N = |f_1 - f_2| = |50 - 46| = 4 \text{ Hz} \]

5. **Determine the number of maximum intensities heard in one second**:
Since the beat frequency is 4 Hz, it means that the observer hears 4 maximum intensities (or beats) per second.

### Final Answer:
The observer hears maximum intensity **4 times in one second**.

---