The equation of a transverse wave is given by `y= 10sin2Π(2x-3t)` where x and y are in cm and t is in s. Its frequency is

Options

A. 1 Hz
B. 4 Hz
C. 2 Hz
D. 3 Hz

Answer

To find the frequency of the transverse wave given by the equation \( y = 10 \sin(2\pi(2x - 3t)) \), we can follow these steps:

### Step 1: Identify the wave equation format
The general form of a transverse wave can be expressed as:
\[ y = A \sin(kx - \omega t) \]
where:
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

### Step 2: Compare the given equation with the general form
From the given equation \( y = 10 \sin(2\pi(2x - 3t)) \), we can rewrite it as:
\[ y = 10 \sin(2\pi \cdot 2x - 2\pi \cdot 3t) \]
This allows us to identify:
- Amplitude \( A = 10 \)
- Wave number \( k = 2\pi \cdot 2 \)
- Angular frequency \( \omega = 2\pi \cdot 3 \)

### Step 3: Calculate the angular frequency
From the comparison, we have:
\[ \omega = 2\pi \cdot 3 \]
Thus, we can simplify:
\[ \omega = 6\pi \, \text{rad/s} \]

### Step 4: Relate angular frequency to frequency
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:
\[ \omega = 2\pi f \]
Substituting the value of \( \omega \):
\[ 6\pi = 2\pi f \]

### Step 5: Solve for frequency \( f \)
Dividing both sides by \( 2\pi \):
\[ f = \frac{6\pi}{2\pi} = 3 \, \text{Hz} \]

### Conclusion
The frequency of the wave is:
\[ \boxed{3 \, \text{Hz}} \]