The following equation represents standing wave set up in a medium ,
`y = 4 cos (pi x)/(3) sin 40 pi t`
where `x and y` are in cm and t in second. Find out the amplitude and the velocity of the two component waves and calculate the distance adjacent nodes . What is the velocity of a medium particle at ` x = 3 cm` at time `1//8 s`?

Explanation

To solve the problem step by step, we will analyze the given standing wave equation and extract the required information.

### Given Equation:
The standing wave is represented by the equation:
\[ y = 4 \cos\left(\frac{\pi x}{3}\right) \sin(40 \pi t) \]

### Step 1: Identify the Amplitude of the Component Waves
The amplitude of the standing wave can be determined from the equation. The amplitude \( A \) of the standing wave is the coefficient of the sine function.

**Solution:**
- The amplitude \( A \) is given as \( 4 \) cm.

### Step 2: Determine the Velocity of the Component Waves
The angular frequency \( \omega \) and the wave number \( k \) can be extracted from the equation.

- From the sine term, we have:
\[ \omega = 40 \pi \]
- From the cosine term, we have:
\[ k = \frac{\pi}{3} \]

The velocity \( v \) of the waves can be calculated using the formula:
\[ v = \frac{\omega}{k} \]

**Solution:**
\[ v = \frac{40 \pi}{\frac{\pi}{3}} = 40 \times 3 = 120 \text{ cm/s} \]

### Step 3: Calculate the Distance Between Adjacent Nodes
The distance between adjacent nodes in a standing wave is given by:
\[ \text{Distance} = \frac{\lambda}{2} \]
where \( \lambda \) is the wavelength.

The wavelength \( \lambda \) can be calculated using the wave number:
\[ k = \frac{2\pi}{\lambda} \Rightarrow \lambda = \frac{2\pi}{k} = \frac{2\pi}{\frac{\pi}{3}} = 6 \text{ cm} \]

**Solution:**
\[ \text{Distance between adjacent nodes} = \frac{6}{2} = 3 \text{ cm} \]

### Step 4: Calculate the Velocity of a Medium Particle at \( x = 3 \) cm and \( t = \frac{1}{8} \) s
The velocity of a medium particle can be found by taking the partial derivative of \( y \) with respect to \( t \):
\[ v_y = \frac{dy}{dt} \]

**Differentiation:**
Using the product rule:
\[ \frac{dy}{dt} = 4 \cos\left(\frac{\pi x}{3}\right) \cdot \frac{d}{dt}[\sin(40 \pi t)] \]
\[ = 4 \cos\left(\frac{\pi x}{3}\right) \cdot (40 \pi \cos(40 \pi t)) \]

Now substituting \( x = 3 \) cm and \( t = \frac{1}{8} \) s:
- First, calculate \( \cos\left(\frac{\pi \cdot 3}{3}\right) = \cos(\pi) = -1 \)
- Next, calculate \( \cos\left(40 \pi \cdot \frac{1}{8}\right) = \cos(5\pi) = -1 \)

**Solution:**
\[ v_y = 4 \cdot (-1) \cdot (40 \pi) \cdot (-1) = 160 \pi \text{ cm/s} \]

### Final Answers:
1. Amplitude: \( 4 \) cm
2. Velocity of component waves: \( 120 \) cm/s
3. Distance between adjacent nodes: \( 3 \) cm
4. Velocity of medium particle at \( x = 3 \) cm and \( t = \frac{1}{8} \) s: \( 160 \pi \) cm/s