The equation of a wave is
`y=(x,t)=0.05 sin [(pi)/(2)(10x-40t)-(pi)/(4)]m`
find: (a) the wavelength, the frequency and the wave velocity
(b) the participle velocity and acceleration at `x=0.5m`and `t = 0.05s`.

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as specified in the question.

### Part (a): Finding Wavelength, Frequency, and Wave Velocity

1. **Identify the wave equation**:
The given wave equation is:
\[ y(x, t) = 0.05 \sin\left(\frac{\pi}{2}(10x - 40t) - \frac{\pi}{4}\right) \text{ m} \]
This can be rewritten as:
\[ y(x, t) = 0.05 \sin(5\pi x - 20\pi t - \frac{\pi}{4}) \]

2. **Extract wave number (k) and angular frequency (ω)**:
From the standard wave equation \(y(x, t) = A \sin(kx - \omega t + \phi)\), we can identify:
- Wave number \(k = 5\pi\)
- Angular frequency \(\omega = 20\pi\)

3. **Calculate the wavelength (λ)**:
The relationship between wave number and wavelength is given by:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{5\pi} = \frac{2}{5} = 0.4 \text{ m} \]

4. **Calculate the frequency (f)**:
The relationship between angular frequency and frequency is:
\[ \omega = 2\pi f \]
Rearranging gives:
\[ f = \frac{\omega}{2\pi} = \frac{20\pi}{2\pi} = 10 \text{ Hz} \]

5. **Calculate the wave velocity (v)**:
The wave velocity can be calculated using:
\[ v = f \lambda = 10 \times 0.4 = 4 \text{ m/s} \]

### Summary of Part (a):
- Wavelength (λ) = 0.4 m
- Frequency (f) = 10 Hz
- Wave velocity (v) = 4 m/s

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### Part (b): Finding Particle Velocity and Acceleration

1. **Calculate the particle velocity (v_p)**:
The particle velocity is given by the time derivative of displacement:
\[ v_p = \frac{\partial y}{\partial t} \]
Differentiating the wave equation:
\[ v_p = 0.05 \cdot (-20\pi) \cos(5\pi x - 20\pi t - \frac{\pi}{4}) \]

2. **Substitute \(x = 0.5\) m and \(t = 0.05\) s**:
First, calculate the argument of the cosine:
\[ 5\pi(0.5) - 20\pi(0.05) - \frac{\pi}{4} = \frac{5\pi}{2} - \pi - \frac{\pi}{4} = \frac{5\pi}{2} - \frac{4\pi}{4} = \frac{5\pi}{2} - \frac{\pi}{4} = \frac{10\pi - \pi}{4} = \frac{9\pi}{4} \]
Now substitute this back into the particle velocity equation:
\[ v_p = -20\pi(0.05) \cos\left(\frac{9\pi}{4}\right) = -1\pi \cos\left(\frac{9\pi}{4}\right) \]
Since \(\cos\left(\frac{9\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\):
\[ v_p = -\pi \cdot \frac{\sqrt{2}}{2} \approx -2.22 \text{ m/s} \]

3. **Calculate the particle acceleration (a_p)**:
The particle acceleration is given by the second time derivative of displacement:
\[ a_p = \frac{\partial^2 y}{\partial t^2} \]
Differentiating the particle velocity:
\[ a_p = 0.05 \cdot (-20\pi)(-20\pi) \sin(5\pi x - 20\pi t - \frac{\pi}{4}) = 400\pi^2(0.05) \sin(5\pi x - 20\pi t - \frac{\pi}{4}) \]
Substitute the same argument into the sine function:
\[ a_p = 20\pi^2 \sin\left(\frac{9\pi}{4}\right) = 20\pi^2 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2}\pi^2 \approx 140 \text{ m/s}^2 \]

### Summary of Part (b):
- Particle velocity (v_p) at \(x = 0.5\) m and \(t = 0.05\) s = \(-2.22\) m/s
- Particle acceleration (a_p) at \(x = 0.5\) m and \(t = 0.05\) s = \(140\) m/s²

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