The equation of a wave travelling on a string is `y=(0.10mm)sin[3.14m^-1)x+(314s^-1)t]`. (a) In which direction does the wave travel ? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string ?

Explanation

Let's solve the problem step by step.

### Given:
The equation of the wave is:
\[ y = (0.10 \, \text{mm}) \sin[3.14 \, \text{m}^{-1} \cdot x + (314 \, \text{s}^{-1}) \cdot t] \]

### (a) Direction of Wave Travel
To determine the direction of wave travel, we need to analyze the sine function in the wave equation. The general form of a wave traveling in the positive x-direction is:
\[ y = a \sin(\omega t - kx) \]
And for a wave traveling in the negative x-direction, it is:
\[ y = a \sin(\omega t + kx) \]

In our case, we have:
\[ y = (0.10 \, \text{mm}) \sin[3.14 \, \text{m}^{-1} \cdot x + (314 \, \text{s}^{-1}) \cdot t] \]
This indicates that the wave is traveling in the **negative x-direction** because of the "+" sign in front of \( kx \).

### (b) Wave Speed, Wavelength, and Frequency
1. **Wave Speed (v)**:
The wave speed can be calculated using the formula:
\[ v = f \lambda \]
where \( f \) is the frequency and \( \lambda \) is the wavelength.

2. **Frequency (f)**:
From the equation, we know:
\[ \omega = 314 \, \text{s}^{-1} \]
The relationship between angular frequency (\( \omega \)) and frequency (\( f \)) is:
\[ \omega = 2\pi f \]
Therefore:
\[ f = \frac{\omega}{2\pi} = \frac{314}{2\pi} \approx 50 \, \text{Hz} \]

3. **Wavelength (\( \lambda \))**:
The wave number \( k \) is given as:
\[ k = 3.14 \, \text{m}^{-1} \]
The relationship between wave number and wavelength is:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{3.14} \approx 2 \, \text{m} \]

4. **Wave Speed Calculation**:
Now substituting \( f \) and \( \lambda \) into the wave speed formula:
\[ v = f \lambda = 50 \, \text{Hz} \times 2 \, \text{m} = 100 \, \text{m/s} \]

### (c) Maximum Displacement and Maximum Speed
1. **Maximum Displacement**:
The maximum displacement (amplitude) is given directly from the wave equation:
\[ A = 0.10 \, \text{mm} = 0.10 \times 10^{-3} \, \text{m} = 0.0001 \, \text{m} \]

2. **Maximum Speed**:
The maximum speed of a point on the string is given by:
\[ v_{max} = \omega A \]
Substituting the values:
\[ v_{max} = 314 \, \text{s}^{-1} \times 0.0001 \, \text{m} = 0.0314 \, \text{m/s} \]

### Final Answers:
- (a) The wave travels in the **negative x-direction**.
- (b) The wave speed is **100 m/s**, the wavelength is **2 m**, and the frequency is **50 Hz**.
- (c) The maximum displacement is **0.10 mm** and the maximum speed is **0.0314 m/s**.