A string vibrate according to the equation `y = 5 sin ((pi x)/(3)) cos (40 pi t)` where x and y in cm's and t is in second.
(i) What is the equation of incident and reflected wave ?
(ii) What is the distance between the adjacent nodes ?
(iii) What is the velocity of the particle of the string at the position x = 1.5 cm when `t = (9)/(8) sec` ?

Explanation

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

### Given:
The equation of the string vibration is:
\[ y = 5 \sin\left(\frac{\pi x}{3}\right) \cos(40 \pi t) \]
where \( x \) and \( y \) are in centimeters and \( t \) is in seconds.

### (i) Equation of Incident and Reflected Wave

1. **Identify the Form of the Standing Wave**:
The given equation is in the form of a standing wave:
\[ y = 2A \sin(kx) \cos(\omega t) \]
Here, \( A = 2.5 \), \( k = \frac{\pi}{3} \), and \( \omega = 40 \pi \).

2. **Use the Relationship for Standing Waves**:
A standing wave can be expressed as the sum of an incident wave and a reflected wave:
\[ y = y_{\text{incident}} + y_{\text{reflected}} \]
Using the identity for sine, we can express the standing wave as:
\[ y = A \sin(kx + \omega t) + A \sin(kx - \omega t) \]

3. **Write the Equations**:
- The equation of the incident wave is:
\[ y_{\text{incident}} = 2.5 \sin\left(\frac{\pi x}{3} + 40 \pi t\right) \]
- The equation of the reflected wave is:
\[ y_{\text{reflected}} = 2.5 \sin\left(\frac{\pi x}{3} - 40 \pi t\right) \]

### (ii) Distance Between Adjacent Nodes

1. **Identify the Wavelength**:
The distance between adjacent nodes in a standing wave is half the wavelength (\(\lambda/2\)). We need to find \(\lambda\).

2. **Calculate Wavenumber**:
The wavenumber \( k \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
From the equation, we have \( k = \frac{\pi}{3} \).

3. **Find the Wavelength**:
Setting the two equations equal:
\[ \frac{\pi}{3} = \frac{2\pi}{\lambda} \]
Solving for \(\lambda\):
\[ \lambda = 6 \text{ cm} \]

4. **Calculate Distance Between Nodes**:
The distance between adjacent nodes is:
\[ \text{Distance} = \frac{\lambda}{2} = \frac{6}{2} = 3 \text{ cm} \]

### (iii) Velocity of the Particle at \( x = 1.5 \) cm and \( t = \frac{9}{8} \) sec

1. **Find the Displacement Equation**:
The displacement at any point is given by:
\[ y = 5 \sin\left(\frac{\pi x}{3}\right) \cos(40 \pi t) \]

2. **Differentiate to Find Velocity**:
The velocity \( v \) of the particle is the time derivative of displacement:
\[ v = \frac{dy}{dt} = 5 \sin\left(\frac{\pi x}{3}\right) \frac{d}{dt}(\cos(40 \pi t)) \]
\[ = 5 \sin\left(\frac{\pi x}{3}\right) \left(-40 \pi \sin(40 \pi t)\right) \]

3. **Substitute \( x = 1.5 \) cm and \( t = \frac{9}{8} \) sec**:
\[ v = -200 \pi \sin\left(\frac{\pi \cdot 1.5}{3}\right) \sin\left(40 \pi \cdot \frac{9}{8}\right) \]

4. **Calculate Each Sine Term**:
- First sine term:
\[ \sin\left(\frac{\pi \cdot 1.5}{3}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
- Second sine term:
\[ 40 \pi \cdot \frac{9}{8} = 45 \pi \]
\[ \sin(45 \pi) = 0 \]

5. **Final Velocity Calculation**:
Since the second sine term is zero:
\[ v = -200 \pi \cdot 1 \cdot 0 = 0 \]

### Summary of Answers
(i) Incident wave: \( y_{\text{incident}} = 2.5 \sin\left(\frac{\pi x}{3} + 40 \pi t\right) \)
Reflected wave: \( y_{\text{reflected}} = 2.5 \sin\left(\frac{\pi x}{3} - 40 \pi t\right) \)
(ii) Distance between adjacent nodes: \( 3 \text{ cm} \)
(iii) Velocity of the particle at \( x = 1.5 \) cm and \( t = \frac{9}{8} \) sec: \( 0 \text{ cm/s} \)