Two harmonic waves are represented in SI units by
`y_1(x,t) = 0.2 sin (x-3.0t) and y_2(x,t) = 0.2 sin (x- 3.0t + phi)`
(a) Write the expression for the sum `y=y_1 + y_2 for phi = pi/2 rad.`
(b) Suppose the phase difference `phi` between the waves is unknown and the amplitude of their sum is 0.32 m, what is `phi`?

Explanation

To solve the problem step by step, we will address each part of the question separately.

### Part (a): Expression for the sum \( y = y_1 + y_2 \) for \( \phi = \frac{\pi}{2} \) rad.

1. **Identify the given equations:**
- \( y_1(x,t) = 0.2 \sin(x - 3t) \)
- \( y_2(x,t) = 0.2 \sin(x - 3t + \phi) \)
- Here, \( \phi = \frac{\pi}{2} \).

2. **Substitute \( \phi \) into \( y_2 \):**
- \( y_2(x,t) = 0.2 \sin(x - 3t + \frac{\pi}{2}) \)

3. **Use the sine addition formula:**
- The sine addition formula states that \( \sin(a + b) = \sin a \cos b + \cos a \sin b \).
- Thus, \( y_2(x,t) = 0.2 \left( \sin(x - 3t) \cos\left(\frac{\pi}{2}\right) + \cos(x - 3t) \sin\left(\frac{\pi}{2}\right) \right) \).
- Since \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \), we have:
- \( y_2(x,t) = 0.2 \cdot \cos(x - 3t) \).

4. **Now, sum \( y_1 \) and \( y_2 \):**
- \( y = y_1 + y_2 = 0.2 \sin(x - 3t) + 0.2 \cos(x - 3t) \).

5. **Factor out the common amplitude:**
- We can express this as:
- \( y = 0.2 \left( \sin(x - 3t) + \cos(x - 3t) \right) \).

6. **Use the amplitude formula:**
- The amplitude of the resultant wave can be calculated using the formula:
- \( A = \sqrt{A_1^2 + A_2^2} \) where \( A_1 = 0.2 \) and \( A_2 = 0.2 \).
- Thus, \( A = \sqrt{(0.2)^2 + (0.2)^2} = \sqrt{0.04 + 0.04} = \sqrt{0.08} = 0.2\sqrt{2} \).

7. **Determine the phase angle \( \theta \):**
- The phase angle \( \theta \) can be found using:
- \( \tan(\theta) = \frac{A_2}{A_1} = \frac{0.2}{0.2} = 1 \).
- Therefore, \( \theta = 45^\circ = \frac{\pi}{4} \) radians.

8. **Final expression for the sum:**
- Thus, the expression for the sum is:
- \( y = 0.2\sqrt{2} \sin\left(x - 3t + \frac{\pi}{4}\right) \).

### Part (b): Find \( \phi \) when the amplitude of the sum is 0.32 m.

1. **Use the amplitude formula:**
- The formula for the amplitude of the sum of two waves is:
- \( A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \).
- Here, \( A = 0.32 \), \( A_1 = 0.2 \), and \( A_2 = 0.2 \).

2. **Substitute the values into the equation:**
- \( 0.32 = \sqrt{(0.2)^2 + (0.2)^2 + 2 \cdot 0.2 \cdot 0.2 \cdot \cos(\phi)} \).

3. **Square both sides:**
- \( (0.32)^2 = (0.2)^2 + (0.2)^2 + 2 \cdot 0.2 \cdot 0.2 \cdot \cos(\phi) \).
- \( 0.1024 = 0.04 + 0.04 + 0.08 \cos(\phi) \).

4. **Simplify the equation:**
- \( 0.1024 = 0.08 + 0.08 \cos(\phi) \).
- \( 0.1024 - 0.08 = 0.08 \cos(\phi) \).
- \( 0.0224 = 0.08 \cos(\phi) \).

5. **Solve for \( \cos(\phi) \):**
- \( \cos(\phi) = \frac{0.0224}{0.08} = 0.28 \).

6. **Find \( \phi \):**
- \( \phi = \cos^{-1}(0.28) \).
- Using a calculator, we find \( \phi \approx 1.29 \) radians.

### Final Answers:
- (a) \( y = 0.2\sqrt{2} \sin\left(x - 3t + \frac{\pi}{4}\right) \)
- (b) \( \phi \approx \pm 1.29 \) radians