Two sinusoidal waves combining in a medium are described by the equations
`y_1 = (3.0 cm) sin pi (x+ 0.60t)`
and `y_2 = (3.0 cm) sin pi (x-0.06 t)`
where, x is in centimetres and t is in seconds. Determine the maximum displacement of the medium at
(a)x=0.250 cm,
(b)x=0.500 cm and
(c) x=1.50 cm.
(d) Find the three smallest values of x corresponding to antinodes.

Explanation

To solve the problem, we will follow these steps systematically:

### Step 1: Write the equations of the waves
The two sinusoidal waves are given by:
- \( y_1 = (3.0 \, \text{cm}) \sin(\pi (x + 0.60t)) \)
- \( y_2 = (3.0 \, \text{cm}) \sin(\pi (x - 0.06t)) \)

### Step 2: Find the resultant wave
The resultant displacement \( y \) of the medium is given by the superposition of the two waves:
\[ y = y_1 + y_2 = 3 \sin(\pi (x + 0.60t)) + 3 \sin(\pi (x - 0.06t)) \]

Using the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
Let \( A = \pi (x + 0.60t) \) and \( B = \pi (x - 0.06t) \).

### Step 3: Calculate \( A + B \) and \( A - B \)
\[ A + B = \pi (x + 0.60t) + \pi (x - 0.06t) = 2\pi x + \pi(0.60t - 0.06t) = 2\pi x + 0.54\pi t \]
\[ A - B = \pi (x + 0.60t) - \pi (x - 0.06t) = \pi(0.60t + 0.06t) = 0.66\pi t \]

### Step 4: Substitute into the resultant wave equation
Now substituting back into the sine addition formula:
\[ y = 3 \cdot 2 \sin\left(\frac{2\pi x + 0.54\pi t}{2}\right) \cos\left(\frac{0.66\pi t}{2}\right) \]
\[ y = 6 \sin(\pi x + 0.27\pi t) \cos(0.33\pi t) \]

### Step 5: Determine maximum displacement
The maximum displacement occurs when \( \cos(0.33\pi t) = 1 \):
\[ y_{\text{max}} = 6 \sin(\pi x) \]

### Step 6: Calculate maximum displacement for given values of x
(a) For \( x = 0.250 \, \text{cm} \):
\[ y_{\text{max}} = 6 \sin(\pi \cdot 0.250) = 6 \sin\left(\frac{\pi}{4}\right) = 6 \cdot \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} \approx 4.24 \, \text{cm} \]

(b) For \( x = 0.500 \, \text{cm} \):
\[ y_{\text{max}} = 6 \sin(\pi \cdot 0.500) = 6 \sin\left(\frac{\pi}{2}\right) = 6 \cdot 1 = 6 \, \text{cm} \]

(c) For \( x = 1.50 \, \text{cm} \):
\[ y_{\text{max}} = 6 \sin(\pi \cdot 1.50) = 6 \sin\left(\frac{3\pi}{2}\right) = 6 \cdot (-1) = -6 \, \text{cm} \]

### Step 7: Find the smallest values of x corresponding to antinodes
Antinodes occur when \( \sin(\pi x) = \pm 1 \):
\[ \pi x = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad x = \frac{1}{2} + n \]
The smallest values of \( x \) for \( n = 0, 1, 2 \) are:
- For \( n = 0 \): \( x = 0.5 \, \text{cm} \)
- For \( n = 1 \): \( x = 1.5 \, \text{cm} \)
- For \( n = 2 \): \( x = 2.5 \, \text{cm} \)

### Final Answers
(a) \( 4.24 \, \text{cm} \)
(b) \( 6 \, \text{cm} \)
(c) \( -6 \, \text{cm} \)
(d) \( 0.5 \, \text{cm}, 1.5 \, \text{cm}, 2.5 \, \text{cm} \)