The vibration of a string of length `60 cm` is represented by the equation,
`y = 3 cos (pix//20) cos(72pit)` where `x` & `y` are in `cm` and `t` in sec.
(i) Write down the compoent waves whose superposition gives the above wave.
(ii) Where are the nodes and antinodes located along the string.
(iii) What is the velocity of the particle of the string at the position `x = 5 cm` & `t = 0.25` sec.

Answer

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

### Given:
The wave equation is:
\[ y = 3 \cos\left(\frac{\pi x}{20}\right) \cos(72 \pi t) \]
where \( x \) and \( y \) are in cm, and \( t \) is in seconds.

### (i) Component Waves
To find the component waves whose superposition gives the above wave, we can use the trigonometric identity:
\[ \cos A \cos B = \frac{1}{2} \left( \cos(A - B) + \cos(A + B) \right) \]

Let:
- \( A = \frac{\pi x}{20} \)
- \( B = 72 \pi t \)

Using the identity, we can rewrite the wave equation:
\[ y = 3 \cos\left(\frac{\pi x}{20}\right) \cos(72 \pi t) = \frac{3}{2} \left( \cos\left(\frac{\pi x}{20} - 72 \pi t\right) + \cos\left(\frac{\pi x}{20} + 72 \pi t\right) \right) \]

Thus, the component waves are:
1. \( y_1 = 1.5 \cos\left(\frac{\pi x}{20} - 72 \pi t\right) \)
2. \( y_2 = 1.5 \cos\left(\frac{\pi x}{20} + 72 \pi t\right) \)

### (ii) Nodes and Antinodes
To find the positions of nodes and antinodes along the string, we need to analyze the wave function.

1. **Nodes** occur where the amplitude of the wave is zero. This happens when:
\[ \cos\left(\frac{\pi x}{20}\right) = 0 \]
The cosine function is zero at:
\[ \frac{\pi x}{20} = \frac{\pi}{2} + n\pi \quad (n = 0, 1, 2, \ldots) \]
Solving for \( x \):
\[ x = 10 + 20n \quad (n = 0, 1, 2, 3) \]
For \( n = 0, 1, 2, 3 \) within the length of the string (60 cm):
- \( n = 0 \): \( x = 10 \) cm
- \( n = 1 \): \( x = 30 \) cm
- \( n = 2 \): \( x = 50 \) cm
- \( n = 3 \): \( x = 70 \) cm (out of bounds)

Thus, nodes are located at:
\[ x = 0, 20, 40, 60 \text{ cm} \]

2. **Antinodes** occur where the amplitude is maximum, which happens when:
\[ \cos\left(\frac{\pi x}{20}\right) = \pm 1 \]
This occurs at:
\[ \frac{\pi x}{20} = n\pi \quad (n = 0, 1, 2, \ldots) \]
Solving for \( x \):
\[ x = 20n \quad (n = 0, 1, 2, 3) \]
For \( n = 0, 1, 2, 3 \):
- \( n = 0 \): \( x = 0 \) cm
- \( n = 1 \): \( x = 20 \) cm
- \( n = 2 \): \( x = 40 \) cm
- \( n = 3 \): \( x = 60 \) cm

Thus, antinodes are located at:
\[ x = 10, 30, 50 \text{ cm} \]

### (iii) Velocity of the Particle at \( x = 5 \) cm and \( t = 0.25 \) sec
To find the velocity of the particle, we need to calculate the partial derivative of \( y \) with respect to \( t \):
\[ v = \frac{\partial y}{\partial t} \]
Given:
\[ y = 3 \cos\left(\frac{\pi x}{20}\right) \cos(72 \pi t) \]
Using the product rule:
\[ v = 3 \cos\left(\frac{\pi x}{20}\right) \frac{\partial}{\partial t} \left( \cos(72 \pi t) \right) \]
Calculating the derivative:
\[ \frac{\partial}{\partial t} \left( \cos(72 \pi t) \right) = -72 \pi \sin(72 \pi t) \]
Thus:
\[ v = -3 \cdot 72 \pi \cos\left(\frac{\pi x}{20}\right) \sin(72 \pi t) \]

Now substituting \( x = 5 \) cm and \( t = 0.25 \) sec:
1. Calculate \( \cos\left(\frac{\pi \cdot 5}{20}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
2. Calculate \( \sin(72 \pi \cdot 0.25) = \sin(18 \pi) = 0 \)

Thus:
\[ v = -3 \cdot 72 \pi \cdot \frac{1}{\sqrt{2}} \cdot 0 = 0 \]

### Final Answers:
(i) Component waves:
- \( y_1 = 1.5 \cos\left(\frac{\pi x}{20} - 72 \pi t\right) \)
- \( y_2 = 1.5 \cos\left(\frac{\pi x}{20} + 72 \pi t\right) \)

(ii) Nodes are located at \( 0, 20, 40, 60 \) cm; Antinodes are located at \( 10, 30, 50 \) cm.

(iii) The velocity of the particle at \( x = 5 \) cm and \( t = 0.25 \) sec is \( 0 \) cm/s.