The maximum particle velocity and acceleration of a harmonically moving transverse wave set up on a string are 5 `ms^(-1) and 100 ms^(-2)` respectively. Write the equation of waveform if velocity of wave is 30 `ms^(-1)`.

Answer

To find the equation of the waveform for the given harmonically moving transverse wave on a string, we can follow these steps:

### Step 1: Identify the given values
- Maximum particle velocity, \( V_{\text{max}} = 5 \, \text{ms}^{-1} \)
- Maximum particle acceleration, \( A_{\text{max}} = 100 \, \text{ms}^{-2} \)
- Velocity of the wave, \( v = 30 \, \text{ms}^{-1} \)

### Step 2: Relate maximum velocity and acceleration to amplitude and angular frequency
The maximum particle velocity is given by:
\[ V_{\text{max}} = A \cdot \omega \]
The maximum particle acceleration is given by:
\[ A_{\text{max}} = A \cdot \omega^2 \]

### Step 3: Set up the equations
From the equations above, we can express \( A \) in terms of \( \omega \):
1. From \( V_{\text{max}} = A \cdot \omega \):
\[ A = \frac{V_{\text{max}}}{\omega} \]
2. From \( A_{\text{max}} = A \cdot \omega^2 \):
\[ A = \frac{A_{\text{max}}}{\omega^2} \]

### Step 4: Equate the two expressions for amplitude
Setting the two expressions for \( A \) equal to each other:
\[ \frac{V_{\text{max}}}{\omega} = \frac{A_{\text{max}}}{\omega^2} \]
Cross-multiplying gives:
\[ V_{\text{max}} \cdot \omega = A_{\text{max}} \]
Substituting the known values:
\[ 5 \cdot \omega = 100 \]
Solving for \( \omega \):
\[ \omega = \frac{100}{5} = 20 \, \text{rad/s} \]

### Step 5: Calculate the amplitude \( A \)
Now, using the value of \( \omega \) to find \( A \):
\[ A = \frac{V_{\text{max}}}{\omega} = \frac{5}{20} = 0.25 \, \text{m} \]

### Step 6: Calculate the wave number \( k \)
Using the relationship between angular frequency, wave speed, and wave number:
\[ \omega = v \cdot k \implies k = \frac{\omega}{v} \]
Substituting the known values:
\[ k = \frac{20}{30} = \frac{2}{3} \, \text{m}^{-1} \]

### Step 7: Write the equation of the wave
The general equation for a wave is given by:
\[ y(x, t) = A \sin(\omega t + kx) \]
Substituting the values of \( A \), \( \omega \), and \( k \):
\[ y(x, t) = 0.25 \sin(20t + \frac{2}{3}x) \]

### Final Equation
Thus, the equation of the waveform is:
\[ y(x, t) = 0.25 \sin(20t + \frac{2}{3}x) \]