A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves the end up and down through a distance by `2.0 cm`. the motion of bar is continuous and is repeated regularly `125` times per second. If klthe distance between adjacent wave crests is observed to be `15.6 cm` and the wave is moving along positive `x-`direction, and at `t=0` the element of the string at `x=0` is at mean position `y =0` and is moving downward, the equation of the wave is best described by

Options

A. `y=(1 cm) sin [(40.3 rad//m) x-(786 rad//s)t]`
B. `y=(2 cm) sin [(40.3 rad//m) x-(786 rad//s)t]`
C. `y=(1 cm) cos [(40.3 rad//m) x-(786 rad//s)t]`
D. `y=(2 cm) cos[(40.3 rad//m) x-(786 rad//s)t]`

Explanation

To find the equation of the transverse sinusoidal wave described in the question, we will follow these steps:

### Step 1: Identify the given parameters
- Amplitude (A): The distance from the mean position to the crest is half the total distance moved by the bar. Given that the bar moves up and down by 2.0 cm, the amplitude \( A = \frac{2.0 \, \text{cm}}{2} = 1.0 \, \text{cm} = 0.01 \, \text{m} \).
- Frequency (f): The bar moves up and down 125 times per second, so \( f = 125 \, \text{Hz} \).
- Wavelength (λ): The distance between adjacent crests is given as 15.6 cm, so \( \lambda = 15.6 \, \text{cm} = 0.156 \, \text{m} \).

### Step 2: Calculate the wave number (k)
The wave number \( k \) is given by the formula:
\[ k = \frac{2\pi}{\lambda} \]
Substituting the value of \( \lambda \):
\[ k = \frac{2\pi}{0.156} \approx 40.24 \, \text{rad/m} \]

### Step 3: Calculate the angular frequency (ω)
The angular frequency \( \omega \) is given by:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \times 125 \approx 785.4 \, \text{rad/s} \]

### Step 4: Determine the direction and phase of the wave
The wave is moving in the positive x-direction, and at \( t = 0 \), the element of the string at \( x = 0 \) is at the mean position \( y = 0 \) and is moving downward. This indicates that the wave function must have a negative sign in front of the angular frequency term to reflect the downward motion at \( t = 0 \).

### Step 5: Write the wave equation
The general form of the wave equation for a wave moving in the positive x-direction is:
\[ y(x, t) = A \sin(kx - \omega t + \phi) \]
Since the wave is at the mean position and moving downward at \( t = 0 \), we can set \( \phi = 0 \) and use the negative sign in front of \( \omega t \):
\[ y(x, t) = A \sin(kx - \omega t) \]
Substituting the values of \( A \), \( k \), and \( \omega \):
\[ y(x, t) = 0.01 \sin(40.24x - 785.4t) \]

### Final Wave Equation
Thus, the equation of the wave is:
\[ y(x, t) = 0.01 \sin(40.24x - 785.4t) \]

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