The equation of a wave travelling on a string is given by Y(mn) = 8 sin[`(5m^(-1)x-(4s^(-1)t`]. Then

Options

A. velocity of wave is 0.8 m/s
B. the displacement of a particle of the sting at t= 0 and `x = (30)` m from the mean position is 4 m
C. the displacement of th mean position at t = 0, `x =(30)`m is 8 m
D. velocity of the wave is 8 m/s

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given wave equation
The wave equation is given as:
\[ Y(mn) = 8 \sin(5x - 4t) \]

### Step 2: Compare with the standard wave equation
The standard form of a wave equation is:
\[ Y(x, t) = A \sin(kx - \omega t) \]
where:
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

From the given equation, we can identify:
- Amplitude \( A = 8 \)
- Wave number \( k = 5 \, \text{m}^{-1} \)
- Angular frequency \( \omega = 4 \, \text{s}^{-1} \)

### Step 3: Calculate the wave velocity
The wave velocity \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]

Substituting the values:
\[ v = \frac{4 \, \text{s}^{-1}}{5 \, \text{m}^{-1}} = 0.8 \, \text{m/s} \]

### Step 4: Calculate the displacement at a specific point and time
We need to find the displacement \( Y \) at \( t = 0 \) and \( x = 30 \, \text{m} \):
\[ Y = 8 \sin(5(30) - 4(0)) \]
\[ Y = 8 \sin(150) \]

### Step 5: Evaluate \( \sin(150) \)
We know that:
\[ \sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin(30^\circ) = \frac{1}{2} \]

### Step 6: Substitute back to find \( Y \)
Now substituting this back:
\[ Y = 8 \cdot \frac{1}{2} = 4 \, \text{m} \]

### Conclusion
1. The wave velocity is \( 0.8 \, \text{m/s} \).
2. The displacement at \( t = 0 \) and \( x = 30 \, \text{m} \) is \( 4 \, \text{m} \).

### Final Answers:
- Velocity: \( 0.8 \, \text{m/s} \)
- Displacement: \( 4 \, \text{m} \)

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