The equation of a transverse wave traveling along a very long string is y 3.0 sin `(0.020pi x -4.0pit),` where x and y are expressed in contimeters and t is in seconds. Determine (a) the amplitude. (b) the wavelength, (c) the frequency (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string (g) What is the transverse displacement at x=3.5 cm when 0.26s ?

Answer

To solve the problem step by step, we will analyze the given wave equation and extract the required parameters.

### Given Wave Equation:
\[ y = 3.0 \sin(0.020\pi x - 4.0\pi t) \]

### Step 1: Determine the Amplitude
The amplitude \( A \) of a wave is the coefficient of the sine function in the wave equation.

**Solution:**
- From the equation, the amplitude \( A = 3.0 \) cm.

### Step 2: Determine the Wavelength
The wave number \( k \) is given by the coefficient of \( x \) in the sine function. The relationship between wave number and wavelength is:
\[ k = \frac{2\pi}{\lambda} \]

**Solution:**
- Here, \( k = 0.020\pi \).
- Setting \( k = 0.020\pi \):
\[ 0.020\pi = \frac{2\pi}{\lambda} \]
- Canceling \( \pi \):
\[ 0.020 = \frac{2}{\lambda} \]
- Rearranging gives:
\[ \lambda = \frac{2}{0.020} = 100 \text{ cm} \]

### Step 3: Determine the Frequency
The angular frequency \( \omega \) is given by the coefficient of \( t \) in the sine function. The relationship between angular frequency and frequency is:
\[ \omega = 2\pi f \]

**Solution:**
- Here, \( \omega = 4.0\pi \).
- Setting \( \omega = 4.0\pi \):
\[ 4.0\pi = 2\pi f \]
- Canceling \( 2\pi \):
\[ f = \frac{4.0}{2} = 2 \text{ Hz} \]

### Step 4: Determine the Speed of the Wave
The speed \( v \) of the wave can be calculated using the formula:
\[ v = f \lambda \]

**Solution:**
- Using \( f = 2 \text{ Hz} \) and \( \lambda = 100 \text{ cm} = 1 \text{ m} \):
\[ v = 2 \times 1 = 2 \text{ m/s} \]

### Step 5: Determine the Direction of Propagation
The direction of propagation can be determined from the signs of the terms in the wave equation. If the term with \( x \) is positive and the term with \( t \) is negative, the wave travels in the positive x-direction.

**Solution:**
- In our equation, the term is \( (0.020\pi x - 4.0\pi t) \), which indicates the wave travels in the positive x-direction.

### Step 6: Determine the Maximum Transverse Speed
The maximum transverse speed \( v_{max} \) of a particle in the string is given by:
\[ v_{max} = \frac{dy}{dt} \]
To find this, we differentiate the wave equation with respect to time \( t \).

**Solution:**
- Differentiating \( y \):
\[ \frac{dy}{dt} = 3.0 \cdot 4\pi \cos(0.020\pi x - 4.0\pi t) \]
- The maximum value of \( \cos \) is 1:
\[ v_{max} = 3.0 \cdot 4\pi = 12\pi \approx 37.7 \text{ cm/s} \]

### Step 7: Determine the Transverse Displacement at \( x = 3.5 \) cm and \( t = 0.26 \) s
We substitute \( x = 3.5 \) cm and \( t = 0.26 \) s into the wave equation to find the transverse displacement \( y \).

**Solution:**
- Substitute into the equation:
\[ y = 3.0 \sin(0.020\pi \cdot 3.5 - 4.0\pi \cdot 0.26) \]
- Calculate:
\[ y = 3.0 \sin(0.070\pi - 1.04\pi) = 3.0 \sin(-0.97\pi) \]
- Since \( \sin(-\theta) = -\sin(\theta) \):
\[ y = 3.0 \cdot (-\sin(0.97\pi)) \approx -0.28 \text{ cm} \]

### Summary of Results:
(a) Amplitude: \( 3.0 \) cm
(b) Wavelength: \( 100 \) cm
(c) Frequency: \( 2 \) Hz
(d) Speed: \( 2 \) m/s
(e) Direction of Propagation: Positive x-direction
(f) Maximum Transverse Speed: \( 37.7 \) cm/s
(g) Transverse Displacement at \( x = 3.5 \) cm and \( t = 0.26 \) s: \( -0.28 \) cm