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)
(3) The number of times `y_(1) + y_(2) = 0` at `x = 0` in `1sec` is

Options

A. 100
B. 46
C. 192
D. 96

Answer

To solve the problem, we need to find the number of times the sum of the two wave functions \( y_1 + y_2 = 0 \) at \( x = 0 \) in 1 second. Let's break this down step by step.

### Step 1: Write down the wave functions
The given wave functions are:
\[ y_1 = A \cos(0.5 \pi x - 100 \pi t) \]
\[ y_2 = A \cos(0.46 \pi x - 92 \pi t) \]

### Step 2: Substitute \( x = 0 \)
We need to find \( y_1 + y_2 \) at \( x = 0 \):
\[ y_1(0, t) = A \cos(-100 \pi t) = A \cos(100 \pi t) \]
\[ y_2(0, t) = A \cos(-92 \pi t) = A \cos(92 \pi t) \]
Thus,
\[ y(0, t) = y_1(0, t) + y_2(0, t) = A \cos(100 \pi t) + A \cos(92 \pi t) \]

### Step 3: Set the equation to zero
We want to find when:
\[ A \cos(100 \pi t) + A \cos(92 \pi t) = 0 \]
Dividing by \( A \) (assuming \( A \neq 0 \)):
\[ \cos(100 \pi t) + \cos(92 \pi t) = 0 \]

### Step 4: Use the cosine addition formula
Using the identity \( \cos A + \cos B = 0 \) can be rewritten as:
\[ \cos(100 \pi t) = -\cos(92 \pi t) \]
This can be expressed as:
\[ \cos(100 \pi t) = \cos\left(\pi - 92 \pi t\right) \]
Thus, we can set the angles equal:
\[ 100 \pi t = (2n + 1)\pi - 92 \pi t \]
where \( n \) is an integer.

### Step 5: Solve for \( t \)
Rearranging gives:
\[ 100 \pi t + 92 \pi t = (2n + 1)\pi \]
\[ 192 \pi t = (2n + 1)\pi \]
Dividing both sides by \( \pi \):
\[ 192 t = 2n + 1 \]
Thus,
\[ t = \frac{2n + 1}{192} \]

### Step 6: Find the number of solutions in 1 second
We need to find the values of \( n \) such that \( t \leq 1 \):
\[ \frac{2n + 1}{192} \leq 1 \]
Multiplying through by 192:
\[ 2n + 1 \leq 192 \]
\[ 2n \leq 191 \]
\[ n \leq 95.5 \]
Since \( n \) must be an integer, the maximum value of \( n \) is 95.

### Step 7: Count the valid integers for \( n \)
The possible values of \( n \) range from \( 0 \) to \( 95 \), which gives us:
\[ n = 0, 1, 2, \ldots, 95 \]
This results in a total of \( 96 \) values.

### Conclusion
The number of times \( y_1 + y_2 = 0 \) at \( x = 0 \) in 1 second is:
\[ \boxed{96} \]