Two waves of equation
`y_(1)=acos(omegat+kx) and y_(2)=acos(omegat-kx)` are superimposed upon each other. They will produce

Options

A. Stationary wave
B. Beats
C. Constructive interference
D. Destructive interference

Answer

To solve the problem of two waves given by the equations \( y_1 = a \cos(\omega t + kx) \) and \( y_2 = a \cos(\omega t - kx) \) superimposing upon each other, we will follow these steps:

### Step 1: Write the resultant wave equation
The resultant wave \( y \) due to the superposition of the two waves is given by:
\[ y = y_1 + y_2 \]
Substituting the equations of the waves:
\[ y = a \cos(\omega t + kx) + a \cos(\omega t - kx) \]

### Step 2: Factor out the common amplitude
We can factor out the amplitude \( a \):
\[ y = a \left( \cos(\omega t + kx) + \cos(\omega t - kx) \right) \]

### Step 3: Use the trigonometric identity
We will use the trigonometric identity for the sum of cosines:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
Here, let \( A = \omega t + kx \) and \( B = \omega t - kx \).

### Step 4: Calculate \( A + B \) and \( A - B \)
Calculating \( A + B \):
\[ A + B = (\omega t + kx) + (\omega t - kx) = 2\omega t \]
Calculating \( A - B \):
\[ A - B = (\omega t + kx) - (\omega t - kx) = 2kx \]

### Step 5: Substitute back into the equation
Now substituting back into the identity:
\[ y = a \left( 2 \cos\left(\frac{2\omega t}{2}\right) \cos\left(\frac{2kx}{2}\right) \right) \]
This simplifies to:
\[ y = 2a \cos(\omega t) \cos(kx) \]

### Step 6: Identify the type of wave
The resultant wave equation \( y = 2a \cos(\omega t) \cos(kx) \) is in the form of a standing wave. The general form of a standing wave is:
\[ y = A \cos(\omega t) \sin(kx) \]
Thus, we can conclude that the superposition of the two waves produces a standing wave.

### Final Result
The two waves will produce a standing wave.

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