A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of 3 m/s and `90 m//s^(2)` respectively. Velocity of the wave is `20 m//s`. Find the waveform.

Explanation

To find the waveform of a harmonically moving transverse wave on a string given the maximum particle velocity, maximum particle acceleration, and wave velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Maximum particle velocity, \( V_{max} = 3 \, \text{m/s} \)
- Maximum particle acceleration, \( A_{max} = 90 \, \text{m/s}^2 \)
- Wave velocity, \( v = 20 \, \text{m/s} \)

2. **Use the Relationship for Maximum Velocity:**
The maximum particle velocity in harmonic motion is given by:
\[ V_{max} = A \cdot \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

3. **Use the Relationship for Maximum Acceleration:**
The maximum particle acceleration is given by:
\[ A_{max} = A \cdot \omega^2 \]

4. **Set Up the Equations:**
From the above relationships, we can write:
- Equation (1): \( A \cdot \omega = 3 \)
- Equation (2): \( A \cdot \omega^2 = 90 \)

5. **Divide Equation (2) by Equation (1):**
\[ \frac{A \cdot \omega^2}{A \cdot \omega} = \frac{90}{3} \]
This simplifies to:
\[ \omega = \frac{90}{3} = 30 \, \text{rad/s} \]

6. **Substitute \( \omega \) Back to Find Amplitude \( A \):**
Substitute \( \omega = 30 \) into Equation (1):
\[ A \cdot 30 = 3 \implies A = \frac{3}{30} = 0.1 \, \text{m} \]

7. **Find the Wave Number \( k \):**
The wave number \( k \) is given by:
\[ k = \frac{\omega}{v} \]
Substitute the values:
\[ k = \frac{30}{20} = 1.5 \, \text{m}^{-1} \]

8. **Write the General Form of the Wave:**
The general form of a wave can be expressed as:
\[ y(x, t) = A \sin(\omega t \pm kx) \]
Substitute the values of \( A \), \( \omega \), and \( k \):
\[ y(x, t) = 0.1 \sin(30t \pm 1.5x) \]

### Final Waveform:
Thus, the waveform of the transverse wave on the string is:
\[ y(x, t) = 0.1 \sin(30t - 1.5x) \quad \text{(or with a plus sign)} \]