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. 10
C. 6
D. 8

Explanation

To solve the problem of how many times an observer hears maximum intensity in one second from the two given sound waves, we will follow these steps:

### Step 1: Identify the equations of the sound waves
The two sound waves are given by:
1. \( y_1(x,t) = A \cos(0.5 \pi x - 100 \pi t) \)
2. \( y_2(x,t) = A \cos(0.46 \pi x - 92 \pi t) \)

### Step 2: Determine the angular frequencies and frequencies
From the equations, we can identify the angular frequencies (\( \omega \)) and the wave numbers (\( k \)):
- For \( y_1 \):
- \( \omega_1 = 100 \pi \)
- \( k_1 = 0.5 \pi \)

- For \( y_2 \):
- \( \omega_2 = 92 \pi \)
- \( k_2 = 0.46 \pi \)

Now we can calculate the frequencies (\( f \)):
- \( f_1 = \frac{\omega_1}{2\pi} = \frac{100 \pi}{2 \pi} = 50 \, \text{Hz} \)
- \( f_2 = \frac{\omega_2}{2\pi} = \frac{92 \pi}{2 \pi} = 46 \, \text{Hz} \)

### Step 3: Calculate the beat frequency
The beat frequency (\( f_b \)) is given by the absolute difference of the two frequencies:
\[ f_b = |f_1 - f_2| = |50 \, \text{Hz} - 46 \, \text{Hz}| = 4 \, \text{Hz} \]

### Step 4: Relate beat frequency to maximum intensity
The number of beats produced per second is equal to the beat frequency. Each beat corresponds to one maximum intensity heard. Therefore, the observer hears maximum intensity 4 times per second.

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

---