To solve the problem, we need to find the wave velocity and the particle velocity at the specified values of \( x \) and \( t \) from the given wave equation:
\[ y(x, t) = 0.02 \sin\left(\frac{x}{0.05} + \frac{t}{0.01}\right) \, \text{m} \]
### Step 1: Identify the parameters from the wave equation
The wave equation can be rewritten in the standard form:
\[ y(x, t) = A \sin(kx + \omega t) \]
Where:
- \( A = 0.02 \, \text{m} \) (amplitude)
- \( k = \frac{1}{0.05} \, \text{m}^{-1} \) (wave number)
- \( \omega = \frac{1}{0.01} \, \text{s}^{-1} \) (angular frequency)
### Step 2: Calculate the wave velocity
The wave velocity \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Calculating \( \omega \) and \( k \):
- \( \omega = \frac{1}{0.01} = 100 \, \text{s}^{-1} \)
- \( k = \frac{1}{0.05} = 20 \, \text{m}^{-1} \)
Now substituting these values into the wave velocity formula:
\[ v = \frac{100}{20} = 5 \, \text{m/s} \]
### Step 3: Calculate the particle velocity
The particle velocity \( v_p \) is given by the time derivative of \( y(x, t) \):
\[ v_p = \frac{\partial y}{\partial t} \]
Calculating the derivative:
\[ v_p = 0.02 \cdot \omega \cdot \cos(kx + \omega t) \]
Substituting \( \omega \) and \( k \):
\[ v_p = 0.02 \cdot 100 \cdot \cos\left(\frac{x}{0.05} + \frac{t}{0.01}\right) \]
### Step 4: Substitute \( x = 0.2 \, \text{m} \) and \( t = 0.3 \, \text{s} \)
Now we need to calculate \( v_p \) at \( x = 0.2 \, \text{m} \) and \( t = 0.3 \, \text{s} \):
1. Calculate \( kx + \omega t \):
\[
kx + \omega t = \frac{0.2}{0.05} + \frac{0.3}{0.01} = 4 + 30 = 34 \, \text{rad}
\]
2. Now substitute this into the particle velocity equation:
\[
v_p = 0.02 \cdot 100 \cdot \cos(34)
\]
Given \( \cos(34) = -0.85 \):
\[
v_p = 0.02 \cdot 100 \cdot (-0.85) = -1.7 \, \text{m/s}
\]
### Final Answers
(a) The wave velocity \( v = 5 \, \text{m/s} \)
(b) The particle velocity at \( x = 0.2 \, \text{m} \) and \( t = 0.3 \, \text{s} \) is \( v_p = -1.7 \, \text{m/s} \)
