A transverse periodic wave is established on a string. The wave is described by the expression
`y= 0.005 sin(20.0x - 2.4pift)`
where y is in meters when x and t are in meters and seconds, respectively. If the wave travels with a speed of 20.0 m/s, what is its frequency: f?

Options

A. 0.16 Hz
B. 0.64 Hz
C. 31.9 Hz
D. 63.7 Hz.

Answer

To find the frequency \( f \) of the wave described by the equation \( y = 0.005 \sin(20.0x - 2.4\pi ft) \), we can follow these steps:

### Step 1: Identify the wave parameters from the equation
The wave equation is given in the form:
\[ y = A \sin(kx - \omega t) \]
where:
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

From the given equation \( y = 0.005 \sin(20.0x - 2.4\pi ft) \), we can identify:
- \( k = 20.0 \) (wave number),
- \( \omega = 2.4\pi f \) (angular frequency).

### Step 2: Relate wave speed, wave number, and angular frequency
The speed \( v \) of the wave is related to the wave number \( k \) and angular frequency \( \omega \) by the formula:
\[ v = \frac{\omega}{k} \]

### Step 3: Substitute the known values
We know:
- The speed of the wave \( v = 20.0 \, \text{m/s} \),
- The wave number \( k = 20.0 \).

Substituting these values into the equation gives:
\[ 20.0 = \frac{2.4\pi f}{20.0} \]

### Step 4: Solve for frequency \( f \)
Rearranging the equation to solve for \( f \):
\[ 20.0 \times 20.0 = 2.4\pi f \]
\[ 400 = 2.4\pi f \]
\[ f = \frac{400}{2.4\pi} \]

### Step 5: Calculate the value of \( f \)
Now, using \( \pi \approx 3.14 \):
\[ f = \frac{400}{2.4 \times 3.14} \approx \frac{400}{7.536} \approx 53.06 \, \text{Hz} \]

### Step 6: Final calculation
To find a more precise value:
\[ f \approx \frac{400}{7.53982} \approx 53.05 \, \text{Hz} \]

### Conclusion
Thus, the frequency \( f \) of the wave is approximately \( 53.05 \, \text{Hz} \).

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