For the wave y= 3.0 sin(36t + 0.018x + pi/4) plot the displacement `(y)` versus `(t)` graphs for `x = 0,`. What is the shaps of the graph ? In which aspects does the oscillatory motion in travelling wave differ from one point to another amplitude, frequency or phase?

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the parameters of the wave equation
The given wave equation is:
\[ y = 3.0 \sin(36t + 0.018x + \frac{\pi}{4}) \]

Here, we can identify:
- Amplitude \( A = 3.0 \)
- Angular frequency \( \omega = 36 \, \text{rad/s} \)
- Phase constant \( \phi = \frac{\pi}{4} \)

### Step 2: Substitute \( x = 0 \) into the wave equation
When \( x = 0 \), the wave equation simplifies to:
\[ y = 3.0 \sin(36t + \frac{\pi}{4}) \]

### Step 3: Calculate the values of \( y \) at different time intervals
To plot the graph, we need to calculate the displacement \( y \) at various time intervals. We will use the period \( T \) to find these intervals.

#### Step 3.1: Calculate the period \( T \)
The relationship between angular frequency \( \omega \) and the period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Thus,
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{36} = \frac{\pi}{18} \, \text{seconds} \]

#### Step 3.2: Choose time intervals
We will choose the following time intervals:
- \( t = 0 \)
- \( t = \frac{T}{8} \)
- \( t = \frac{2T}{8} \)
- \( t = \frac{3T}{8} \)
- \( t = \frac{4T}{8} \)
- \( t = \frac{5T}{8} \)
- \( t = \frac{6T}{8} \)
- \( t = \frac{7T}{8} \)
- \( t = T \)

Calculating \( T/8 \):
\[ \frac{T}{8} = \frac{\pi/18}{8} = \frac{\pi}{144} \]

### Step 4: Calculate \( y \) for each time interval
Now we will calculate \( y \) for each time interval:

1. \( t = 0 \):
\[ y = 3.0 \sin(0 + \frac{\pi}{4}) = 3.0 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} \approx 2.12 \]

2. \( t = \frac{T}{8} \):
\[ y = 3.0 \sin(36 \cdot \frac{\pi}{144} + \frac{\pi}{4}) = 3.0 \sin(\frac{\pi}{2}) = 3.0 \]

3. \( t = \frac{2T}{8} \):
\[ y = 3.0 \sin(36 \cdot \frac{2\pi}{144} + \frac{\pi}{4}) = 3.0 \sin(\frac{3\pi}{4}) = 3.0 \cdot \frac{1}{\sqrt{2}} \approx 2.12 \]

4. \( t = \frac{3T}{8} \):
\[ y = 3.0 \sin(36 \cdot \frac{3\pi}{144} + \frac{\pi}{4}) = 3.0 \sin(\pi) = 0 \]

5. \( t = \frac{4T}{8} \):
\[ y = 3.0 \sin(36 \cdot \frac{4\pi}{144} + \frac{\pi}{4}) = 3.0 \sin(\frac{5\pi}{4}) = -3.0 \cdot \frac{1}{\sqrt{2}} \approx -2.12 \]

6. \( t = \frac{5T}{8} \):
\[ y = 3.0 \sin(36 \cdot \frac{5\pi}{144} + \frac{\pi}{4}) = 3.0 \sin(-\frac{\pi}{2}) = -3.0 \]

7. \( t = \frac{6T}{8} \):
\[ y = 3.0 \sin(36 \cdot \frac{6\pi}{144} + \frac{\pi}{4}) = 3.0 \sin(-\frac{3\pi}{4}) = -3.0 \cdot \frac{1}{\sqrt{2}} \approx -2.12 \]

8. \( t = \frac{7T}{8} \):
\[ y = 3.0 \sin(36 \cdot \frac{7\pi}{144} + \frac{\pi}{4}) = 3.0 \sin(-\pi) = 0 \]

9. \( t = T \):
\[ y = 3.0 \sin(36 \cdot \frac{2\pi}{18} + \frac{\pi}{4}) = 3.0 \sin(\frac{5\pi}{4}) = -3.0 \cdot \frac{1}{\sqrt{2}} \approx -2.12 \]

### Step 5: Plot the graph
Now we can plot the values of \( y \) against \( t \):
- At \( t = 0 \), \( y \approx 2.12 \)
- At \( t = \frac{T}{8} \), \( y = 3.0 \)
- At \( t = \frac{2T}{8} \), \( y \approx 2.12 \)
- At \( t = \frac{3T}{8} \), \( y = 0 \)
- At \( t = \frac{4T}{8} \), \( y \approx -2.12 \)
- At \( t = \frac{5T}{8} \), \( y = -3.0 \)
- At \( t = \frac{6T}{8} \), \( y \approx -2.12 \)
- At \( t = \frac{7T}{8} \), \( y = 0 \)
- At \( t = T \), \( y \approx -2.12 \)

### Final Shape of the Graph
The shape of the graph is sinusoidal, oscillating between the maximum amplitude of \( 3.0 \) and the minimum amplitude of \( -3.0 \).

### Step 6: Analyze the differences in oscillatory motion
In a traveling wave, the oscillatory motion can differ from one point to another in terms of:
- **Phase**: Different points in space may have different phases of oscillation.
- **Amplitude**: In some cases, amplitude can vary with position, but in this case, it remains constant.
- **Frequency**: The frequency remains constant throughout the wave.