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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.

Період плоскої звукової хвилі дорівнює 5 мс, амплітуда 0,2 мм та довжина хвилі 1,5 м. Для точки середовища, що знаходиться на 5 м від джерела, визначити зміщення, швидкість і прискорення в момент часу t = 10 мс.

Learn how to calculate the displacement, velocity, and acceleration of a sound wave with a period of 5 ms, amplitude of 0.2 mm, and wavelength of 1.5 m, at a point 5 meters from the source. This example includes formulas for harmonic wave motion, trigonometric functions, and is suitable for undergraduate physics students.

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