The equation for the transverse wave travelling along a string is y = 4 sin `2pi ((t)/(0.05) -(x)/(60))` lengths expressed in cm and time period in sec. Calculate the wave velocity and maximum particle velocity.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the parameters from the wave equation
The given wave equation is:
\[ y = 4 \sin \left( 2\pi \left(\frac{t}{0.05}\right) - \left(\frac{x}{60}\right) \right) \]

This can be compared to the standard form of a wave equation:
\[ y = A \sin(\omega t - kx) \]

From this comparison, we can identify:
- Amplitude \( A = 4 \) cm
- Angular frequency \( \omega = \frac{2\pi}{0.05} \) rad/s
- Wave number \( k = \frac{2\pi}{60} \) rad/cm

### Step 2: Calculate wave velocity
The wave velocity \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]

Substituting the values of \( \omega \) and \( k \):
\[ v = \frac{\frac{2\pi}{0.05}}{\frac{2\pi}{60}} \]

The \( 2\pi \) terms cancel out:
\[ v = \frac{60}{0.05} \]

Calculating this gives:
\[ v = 1200 \text{ cm/s} = 12 \text{ m/s} \]

### Step 3: Calculate maximum particle velocity
The maximum particle velocity \( v_{max} \) can be found using the formula:
\[ v_{max} = A \omega \]

Substituting the values of \( A \) and \( \omega \):
\[ v_{max} = 4 \cdot \frac{2\pi}{0.05} \]

Calculating this gives:
\[ v_{max} = 4 \cdot \frac{2\pi}{0.05} = \frac{8\pi}{0.05} \]

Calculating \( \frac{8\pi}{0.05} \):
\[ v_{max} = 8\pi \cdot 20 = 160\pi \]

Using \( \pi \approx 3.14 \):
\[ v_{max} \approx 160 \cdot 3.14 \approx 502.4 \text{ cm/s} \]

### Final Answers:
- Wave velocity \( v = 12 \text{ m/s} \)
- Maximum particle velocity \( v_{max} \approx 502.4 \text{ cm/s} \)

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