The transverse wave represented by the equation ` y = 4 "sin"((pi)/(6)) "sin"(3x - 15 t)` has

Options

A. amplitude = `4 pi`
B. wavelength `= 4 pi /3`
C. speed of propagation = 5
D. period `= pi / 15 `

Answer

To analyze the transverse wave represented by the equation \( y = 4 \sin\left(\frac{\pi}{6}\right) \sin(3x - 15t) \), we will extract the relevant parameters and calculate the amplitude, wavelength, speed of propagation, and time period step by step.

### Step 1: Identify the Amplitude
The amplitude \( A \) of the wave is the coefficient of the sine function in the wave equation.

Given:
\[ y = 4 \sin\left(\frac{\pi}{6}\right) \sin(3x - 15t) \]

First, we need to evaluate \( \sin\left(\frac{\pi}{6}\right) \):
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

Now substituting this back into the equation:
\[ y = 4 \cdot \frac{1}{2} \sin(3x - 15t) = 2 \sin(3x - 15t) \]

Thus, the amplitude \( A \) is:
\[ A = 2 \]

### Step 2: Determine the Wave Number \( k \)
The wave number \( k \) is the coefficient of \( x \) in the sine function of the wave equation.

From the equation:
\[ y = 2 \sin(3x - 15t) \]

We see that:
\[ k = 3 \]

### Step 3: Calculate the Wavelength \( \lambda \)
The wavelength \( \lambda \) is related to the wave number \( k \) by the formula:
\[ k = \frac{2\pi}{\lambda} \]

Rearranging gives:
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{3} \]

### Step 4: Find the Angular Frequency \( \omega \)
The angular frequency \( \omega \) is the coefficient of \( t \) in the sine function of the wave equation.

From the equation:
\[ y = 2 \sin(3x - 15t) \]

We see that:
\[ \omega = 15 \]

### Step 5: Calculate the Speed of Propagation \( v \)
The speed of propagation \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]

Substituting the values:
\[ v = \frac{15}{3} = 5 \]

### Step 6: Determine the Time Period \( T \)
The time period \( T \) is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{15} \]

### Summary of Results
- Amplitude \( A = 2 \)
- Wavelength \( \lambda = \frac{2\pi}{3} \)
- Speed of propagation \( v = 5 \)
- Time period \( T = \frac{2\pi}{15} \)