give answer in 2 step with explanation at the end of each step and final answer at the end:Consider the following y-x graph of a wave's position. At
time t = 0, Xo=2.0m, yo=1.2m, and ya =2.0m. v(m)
Fax
Using the graph, construct the expression for the y-position
(xt) of the wave as a function of position x and time ¢ if the Y J I
period of motion is 6.0 s, and the wave moves to the right
(toward the positive x-direction).
x(m)
Use exact rational and irrational coefficients, but express the
phase angle in decimal form rounded two significant figures. \ / x,
Vmax
yan= 2 sin(0 32x — 3)
Incorrect Answer
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:Consider the following y-x graph of a wave's position. At
time t = 0, Xo=2.0m, yo=1.2m, and ya =2.0m. v(m)
Fax
Using the graph, construct the expression for the y-position
(xt) of the wave as a function of position x and time ¢ if the Y J I
period of motion is 6.0 s, and the wave moves to the right
(toward the positive x-direction).
x(m)
Use exact rational and irrational coefficients, but express the
phase angle in decimal form rounded two significant figures. \ / x,
Vmax
yan= 2 sin(0 32x — 3)
Incorrect Answer
This is AI generated Response.
Asked by: Rahul
Created at: 2025-11-19 04:55:11
Final Answer:
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Let's solve the problem step-by-step as requested.
---
## **Step 1: Write the General Wave Equation**
A traveling wave moving to the right can be written as:
\[
y(x, t) = A \sin(kx - \omega t + \phi)
\]
Where:
- \( A \) = amplitude = \( y_{max} = 2. \) m
- \( k \) = wave number = \( \frac{2\pi}{\lambda} \)
- \( \omega \) = angular frequency = \( \frac{2\pi}{T} \)
- \( \phi \) = phase constant
Given:
- Period \( T = 6. \) s
- At \( t = \), \( x_ = 2. \) m, \( y_ = 1.2 \) m
First, calculate \( \omega \):
\[
\omega = \frac{2\pi}{T} = \frac{2\pi}{6.} = \frac{\pi}{3} \ \text{rad/s}
\]
**Explanation:**
We set up the general form and insert values for amplitude and angular frequency based on the problem statement.
---
## **Step 2: Find Wave Number \( k \) and Phase \( \phi \)**
### **Find the wavelength \(\lambda\) and \(k\):**
From the graph, notice the distance between two peaks (full cycle) is the wavelength. The graph shows two peaks from \( x = \) to \( x = 8 \) m, so \( \lambda = 8. \) m.
\[
k = \frac{2\pi}{\lambda} = \frac{2\pi}{8.} = \frac{\pi}{4} \ \text{rad/m}
\]
### **Find phase constant \( \phi \):**
At \( t = \), \( x = x_ = 2. \) m, \( y_ = 1.2 \) m:
\[
y_ = A \sin(kx_ + \phi)
\]
\[
1.2 = 2. \sin\left(\frac{\pi}{4} \cdot 2. + \phi\right)
\]
\[
1.2 = 2. \sin\left(\frac{\pi}{2} + \phi\right)
\]
\[
.6 = \sin\left(\frac{\pi}{2} + \phi\right)
\]
But \(\sin(\frac{\pi}{2} + \phi) = \cos(\phi)\), so:
\[
.6 = \cos(\phi)
\implies \phi = \arccos(.6) \approx .93
\]
**Explanation:**
We use the position at a known \( x_ \) and \( y_ \) to solve for the phase shift \( \phi \).
---
## **Final Answer**
\[
\boxed{
y(x,t) = 2. \sin\left(\frac{\pi}{4} x - \frac{\pi}{3} t + .93 \right)
}
\]
Where:
- \( \frac{\pi}{4} \) is the wave number,
- \( \frac{\pi}{3} \) is the angular frequency,
- \( .93 \) is the phase angle (rounded to two significant figures).
---
**Summary of Steps:**
1. **Wrote the general wave equation and substituted amplitude and frequency.**
2. **Calculated wavelength, wave number, and solved for phase using initial conditions.**
3. **Assembled the final function.**
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