A wave is represented by the equation
`y=A sin314[(t)/(0.5s)-(x)/(100 m)]`
The frequency is n and the wavelength is `lambda`.Then:

Options

A. `n=2 Hz`
B. `n=100 Hz`
C. `lambda=2m`
D. `lambda=100 m`

Explanation

To solve the problem, we need to extract the frequency \( n \) and the wavelength \( \lambda \) from the given wave equation:

### Given Wave Equation:
\[ y = A \sin\left(314 \left(\frac{t}{0.5 \, \text{s}} - \frac{x}{100 \, \text{m}}\right)\right) \]

### Step 1: Identify the Angular Frequency \( \omega \)
The angular frequency \( \omega \) is the coefficient of \( t \) in the wave equation. We can express it as:
\[ \omega = 314 \, \text{(from the equation)} \]

### Step 2: Calculate the Frequency \( n \)
The relationship between angular frequency \( \omega \) and frequency \( n \) is given by:
\[ \omega = 2\pi n \]

We can rearrange this to find \( n \):
\[ n = \frac{\omega}{2\pi} \]

Substituting the value of \( \omega \):
\[ n = \frac{314}{2\pi} \]

Calculating this value:
\[ n \approx \frac{314}{6.28} \approx 50 \, \text{Hz} \]

### Step 3: Identify the Wave Number \( k \)
The wave number \( k \) is the coefficient of \( x \) in the wave equation. It can be expressed as:
\[ k = \frac{314}{100} \]

### Step 4: Calculate the Wavelength \( \lambda \)
The relationship between wave number \( k \) and wavelength \( \lambda \) is given by:
\[ k = \frac{2\pi}{\lambda} \]

Rearranging this gives:
\[ \lambda = \frac{2\pi}{k} \]

Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{\frac{314}{100}} \]

Calculating this value:
\[ \lambda = \frac{2\pi \cdot 100}{314} \approx \frac{628.32}{314} \approx 2 \, \text{m} \]

### Final Results:
- Frequency \( n \approx 50 \, \text{Hz} \)
- Wavelength \( \lambda \approx 2 \, \text{m} \)

### Summary:
The frequency \( n \) is approximately \( 50 \, \text{Hz} \) and the wavelength \( \lambda \) is \( 2 \, \text{m} \).

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