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) :
At x=0 how many times the amplitude of `y_(1) + y_(2)` is zero in one second :-

Options

A. a. 4
B. b. 2
C. c. 1
D. d. 3

Answer

To solve the problem, we need to determine how many times the amplitude of the resultant wave \(y_1 + y_2\) is zero in one second, given the equations of the two sound waves.

### Step 1: Write down the equations of the waves at \(x = 0\)

The equations of the waves are given as:
- \(y_1(x, t) = A \cos(0.5 \pi x - 100 \pi t)\)
- \(y_2(x, t) = A \cos(0.46 \pi x - 92 \pi t)\)

At \(x = 0\), these equations simplify to:
- \(y_1(0, t) = A \cos(-100 \pi t) = A \cos(100 \pi t)\)
- \(y_2(0, t) = A \cos(-92 \pi t) = A \cos(92 \pi t)\)

### Step 2: Write the resultant wave equation

The resultant wave \(y\) can be expressed as:
\[ y = y_1 + y_2 = A \cos(100 \pi t) + A \cos(92 \pi t) \]

### Step 3: Use the cosine addition formula

Using the cosine addition formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
we can rewrite the resultant wave:
- Let \(A = 100 \pi t\) and \(B = 92 \pi t\)

Calculating:
- \(A + B = 100 \pi t + 92 \pi t = 192 \pi t\)
- \(A - B = 100 \pi t - 92 \pi t = 8 \pi t\)

Thus, we have:
\[ y = 2A \cos\left(96 \pi t\right) \cos\left(4 \pi t\right) \]

### Step 4: Determine when the amplitude is zero

The amplitude \(y\) will be zero when either of the cosine terms is zero:
1. \(\cos(96 \pi t) = 0\)
2. \(\cos(4 \pi t) = 0\)

#### For \(\cos(96 \pi t) = 0\):
This occurs when:
\[ 96 \pi t = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Solving for \(t\):
\[ t = \frac{1}{192}(1 + 2n) \quad (n \in \mathbb{Z}) \]
The period of this function is:
\[ T_1 = \frac{1}{48} \text{ seconds} \]
Thus, it crosses zero \(48\) times in one second.

#### For \(\cos(4 \pi t) = 0\):
This occurs when:
\[ 4 \pi t = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Solving for \(t\):
\[ t = \frac{1}{8}(1 + 2n) \quad (n \in \mathbb{Z}) \]
The period of this function is:
\[ T_2 = \frac{1}{4} \text{ seconds} \]
Thus, it crosses zero \(4\) times in one second.

### Step 5: Combine the results

The total number of times the amplitude \(y\) is zero in one second is the sum of the zero crossings from both components:
\[ \text{Total zero crossings} = 48 + 4 = 52 \]

### Final Answer

The amplitude of \(y_1 + y_2\) is zero a total of **52 times in one second**.

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