Waves `y_(1) = Acos(0.5pix - 100pit)` and `y_(2)=Acos(0.46pix - 92pit)` are travelling along x-axis. (Here `x` is in `m` and `t` is in second)
(1) Find the number of times intensity is maximum in time interval of `1 sec`.

Options

A. (a)4
B. (b)6
C. (c)8
D. (d)10

Explanation

To solve the problem, we need to analyze the two wave equations given and determine how many times the intensity is maximum within a time interval of 1 second.

### Step-by-Step Solution:

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

2. **Determine the Frequencies**:
The general form of a wave equation is \( y = A \cos(kx - \omega t) \), where \( \omega = 2\pi f \) (angular frequency) and \( f \) is the frequency.
- For \( y_1 \):
\[ \omega_1 = 100\pi \implies f_1 = \frac{100\pi}{2\pi} = 50 \text{ Hz} \]
- For \( y_2 \):
\[ \omega_2 = 92\pi \implies f_2 = \frac{92\pi}{2\pi} = 46 \text{ Hz} \]

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

4. **Determine the Number of Maxima**:
In one second, the number of beats (or maxima) is equal to the beat frequency. Since one beat consists of one maximum and one minimum:
- The number of times intensity is maximum in 1 second is equal to the beat frequency:
\[ \text{Number of maxima} = f_b = 4 \]

5. **Final Answer**:
Therefore, the number of times the intensity is maximum in the time interval of 1 second is:
\[ \boxed{4} \]