Let's go through this problem step by step.

We are given the following dataLet's go through this problem step by step.

We are given the following

1. Period of the wave (T): $T = 5 \, \text{ms} = 0.005 \, \text{s}$
2. Amplitude of the wave (A): $A = 0.2 \, \text{mm} = 0.0002 \, \text{m}$
3. Wavelength of the wave (λ): $\lambda = 1.5 \, \text{m}$
4. Position of the point: $x = 5 \, \text{m}$
5. Time: $t = 10 \, \text{ms} = 0.01 \, \text{s}$

We need to find:

• Displacement ($\xi(x, t)$)
• Velocity ($v(x, t)$)
• Acceleration ($a(x, t)$)

Step 1: Angular Frequency and Wave Number

First, we calculate the angular frequency ($\omega$) and wave number ($k$).

1. Angular frequency ($\omega$): $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.005} = 400\pi \, \text{rad/s}$

2. Wave number ($k$): $k = \frac{2\pi}{\lambda} = \frac{2\pi}{1.5} = \frac{2\pi}{1.5} \approx 4.19 \, \text{rad/m}$

Step 2: Displacement Function

The displacement of a point in a wave can be described by: $\xi(x, t) = A \cos(kx - \omega t)$

For $x = 5 \, \text{m}$ and $t = 0.01 \, \text{s}$, we substitute the values: $\xi(5, 0.01) = 0.0002 \cos((4.19)(5) - (400\pi)(0.01))$

Let's calculate this step-by-step:

1. Calculate $kx$: $4.19 \times 5 = 20.95$
2. Calculate $\omega t$: $400\pi \times 0.01 \approx 12.57$
3. Substitute these values: $\xi(5, 0.01) = 0.0002 \cos(20.95 - 12.57)$

Calculate $20.95 - 12.57 \approx 8.38$, and then: $\xi(5, 0.01) = 0.0002 \cos(8.38) \approx 0.0002 \times (-0.491) = -0.0000982 \, \text{m}$

Step 3: Velocity Function

The velocity is the time derivative of displacement: $v(x, t) = \frac{\partial \xi(x, t)}{\partial t} = A \omega \sin(kx - \omega t)$

Substitute the values: $v(5, 0.01) = 0.0002 \times 400\pi \times \sin(20.95 - 12.57)$

Using $\sin(8.38) \approx 0.871$: $v(5, 0.01) \approx 0.0002 \times 400\pi \times 0.871 \approx 0.219 \, \text{m/s}$

Step 4: Acceleration Function

The acceleration is the time derivative of velocity: $a(x, t) = -A \omega^2 \cos(kx - \omega t)$

Substitute the values: $a(5, 0.01) = -0.0002 \times (400\pi)^2 \cos(20.95 - 12.57)$

Using $\cos(8.38) \approx -0.491$: $a(5, 0.01) \approx -0.0002 \times (400\pi)^2 \times -0.491 \approx 49.6 \, \text{m/s}^2$

Final Answers

1. Displacement: $\xi(5, 0.01) \approx -0.0000982 \, \text{m}$
2. Velocity: $v(5, 0.01) \approx 0.219 \, \text{m/s}$
3. Acceleration: $a(5, 0.01) \approx 49.6 \, \text{m/s}^2$

Would you like more details on any of these steps?

Here are some related questions for further practice:

1. How would the results change if the amplitude were doubled?
2. What effect does increasing the wavelength have on the displacement at a fixed point?
3. How would the acceleration change if the period of the wave was decreased?
4. Can you derive the velocity expression from the displacement function for a general sinusoidal wave?
5. How would you calculate the energy transported by this wave over time?

Tip: Remember, angular frequency $\omega$ and wave number $k$ are essential in determining the behavior of waves in sinusoidal wave functions.