1. For the case of a transverse wave moving to the right, determine the parameters of period, frequency, wavelength, transverse velocity, the equation of transverse position of the wave, maximum values of transverse velocity and acceleration, knowing that the angular frequency is 6.8 rad/s, the wave number is 4 m-1 and the phase constant is π/4 rad. 2. The values of the position, velocity and acceleration of the movement for x = 4.5m and t = 6sec

**Given:**

Angular frequency, \omega = 6.8 \, \text{rad/s}

Wave number, k = 4 \, \text{m}^{-1}

Phase constant, \phi = \frac{\pi}{4} \, \text{rad}

**Formulas:**

1. Period, T = \frac{2\pi}{\omega}

2. Frequency, f = \frac{1}{T}

3. Wavelength, \lambda = \frac{2\pi}{k}

4. Transverse velocity, v = \omega \cdot A

5. Equation of transverse position of the wave, y(x, t) = A \cdot \cos(kx - \omega t + \phi)

6. Maximum transverse velocity, v_{\text{max}} = \omega \cdot A = 6.8m/s

7. Maximum acceleration, a_{\text{max}} = \omega^2 \cdot A =46.24 m/s^2

**Solution:**

1. Calculate Period:
T=\frac{2\pi}{\omega}=\frac{2\pi}{6.8}\approx0.9239\,\text{s}

2. Calculate Frequency:
f=\frac{1}{T}=\frac{1}{0.9239}\approx1.08\,\text{Hz}

3. Calculate Wavelength:
\lambda = \frac{2\pi}{k} = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{m}

4. Calculate Transverse Velocity:
v = \omega \cdot A

5. Equation of Transverse Position of the Wave:
y(x, t) = A \cdot \cos(kx - \omega t + \phi)

6. Calculate Maximum Transverse Velocity:
v_{\text{max}}=6.8

7. Calculate Maximum Acceleration:
a_{\text{max}}=46.24

**Answers:**

1. Period T = 0.9259 \, \text{s}

2. Frequency f = 1.08 \, \text{Hz}

3. Wavelength \lambda = \frac{\pi}{2} \, \text{m}

4. Transverse velocity v = \omega \cdot A

5. Equation of transverse position of the wave y(x,t)=\cos(4x-6.8t+\frac{\pi}{4})

6. Maximum transverse velocity v_{\text{max}}=6.8

7. Maximum acceleration a_{\text{max}}=46.24