A `6.00m` segment of a long string has a mass of `180 g`. A high-speed photograph shows that the segment contlains four complete cycles of a wave. The string is vibrating sinusoidally with a frequency of `50.0 Hz` and a peak - to - valley dispalcement of `15.0 cm`. (The "peak-to-valley" displacment is the vertical distance from the furthest positive displacement to the farther negative displacement.) (a) Write the funcation that describe this wave traveling in the positive `x` direction.

Explanation

To find the function that describes the wave traveling in the positive x direction, we can follow these steps:

### Step 1: Identify the parameters of the wave
- The mass of the string segment is given as \( m = 180 \, \text{g} = 0.180 \, \text{kg} \).
- The length of the string segment is \( L = 6.00 \, \text{m} \).
- The frequency of the wave is \( f = 50.0 \, \text{Hz} \).
- The peak-to-valley displacement (which is twice the amplitude) is \( 15.0 \, \text{cm} = 0.15 \, \text{m} \).

### Step 2: Calculate the amplitude
The amplitude \( A \) is half of the peak-to-valley displacement:
\[ A = \frac{15.0 \, \text{cm}}{2} = 7.5 \, \text{cm} = 0.075 \, \text{m} \]

### Step 3: Calculate the wave number \( k \)
The wave number \( k \) is given by the formula:
\[ k = \frac{2\pi}{\lambda} \]
First, we need to find the wavelength \( \lambda \). Since there are 4 complete cycles in a length of 6 m:
\[ 4\lambda = 6 \implies \lambda = \frac{6}{4} = 1.5 \, \text{m} \]
Now substituting \( \lambda \) into the equation for \( k \):
\[ k = \frac{2\pi}{1.5} = \frac{4\pi}{3} \, \text{rad/m} \]

### Step 4: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is given by:
\[ \omega = 2\pi f \]
Substituting the frequency:
\[ \omega = 2\pi \times 50.0 = 100\pi \, \text{rad/s} \]

### Step 5: Write the wave function
The general form of the wave traveling in the positive x direction is:
\[ y(x, t) = A \sin(\omega t - kx + \phi) \]
Assuming there is no phase shift (\( \phi = 0 \)), we can write:
\[ y(x, t) = 0.075 \sin(100\pi t - \frac{4\pi}{3} x) \]

### Final Answer
Thus, the function that describes the wave traveling in the positive x direction is:
\[ y(x, t) = 0.075 \sin(100\pi t - \frac{4\pi}{3} x) \]

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