Two plane harmonic sound waves are expressed by the equations
`y_(1)(x,t)=Acos((pi)/(2)x-100pit)`
`y_(2)(x,t)=Acos((46pi)/(100)x-92pit)`
All quantities taken in MKS.
Q. how many times does an observer hear maximum intensity in one second?

Options

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

Answer

To solve the problem, we need to determine how many times an observer hears maximum intensity in one second when two plane harmonic sound waves interfere.

### Step-by-Step Solution:

1. **Identify the given wave equations**:
- The first wave is given by:
\[ y_1(x,t) = A \cos\left(\frac{\pi}{2} x - 100 \pi t\right) \]
- The second wave is given by:
\[ y_2(x,t) = A \cos\left(\frac{46\pi}{100} x - 92 \pi t\right) \]

2. **Extract the angular frequencies**:
- For wave 1, the angular frequency \(\omega_1\) can be identified from the equation:
\[ \omega_1 = 100 \pi \]
- For wave 2, the angular frequency \(\omega_2\) is:
\[ \omega_2 = 92 \pi \]

3. **Convert angular frequencies to frequencies**:
- The frequency \(f\) is related to angular frequency \(\omega\) by the formula:
\[ \omega = 2 \pi f \]
- For wave 1:
\[ 2 \pi f_1 = 100 \pi \implies f_1 = \frac{100}{2} = 50 \text{ Hz} \]
- For wave 2:
\[ 2 \pi f_2 = 92 \pi \implies f_2 = \frac{92}{2} = 46 \text{ Hz} \]

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

5. **Determine the number of times maximum intensity is heard**:
- The beat frequency indicates how many times the observer hears maximum intensity in one second. Since the beat frequency is 4 Hz, the observer hears maximum intensity 4 times in one second.

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

---