Pressure Waves in an Open Tube

In summary, the pressure in an air tube of length L stretched along the x-axis is given byP(x, t) = Patm + P1(x, t) + P2(x, t) (1). The maximum pressure will be at x=0 (where the tube is open). The nearest node will be at t=1/2(period), or 0.00245s.
  • #1
Kara4566
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0
1. Problem Statement:

The pressure in an gas tube of length L stretched along the x-axis is given by
P(x, t) = Patm + P1(x, t) + P2(x, t) (1) where ,

P1(x, t) = 20 [P a] sin (−5.9 x − 1300 t )
P2(x, t) = 20 [P a] sin (5.9 x − 1300 t )One open end of the air tube is at x = 0m. By how much does the pressure increase between the minimum and maximum pressure at that location?

What is the shortest possible air tube which is consistent with this assuming it is closed at x = L?

My attempt at a solution:

The maximum pressure will be at x=0 (where the tube is open). This results in both P1 and P2 equalling zero and therefore the maximum pressure is the atmospheric pressure. Now I attempt to find the minimum pressure and the nearest node. In order to do so, I require the position (x) and the time (t) where this node occurs. To find these variables, I used the equation

frequency= v(n-0.5)/2L, where n=1 (as we are just focusing on the first node).

I calculated length to be 0.27m by the equation λ=2L/(n-0.5), where n=1 and the wavelength is calculated from the first wave equation.

Subbing this back into the equation for frequency, I obtained 203.7 Hz, then found the period to be 0.0049s. Therefore, the node should be located when t=1/2(period), or 0.00245s.

To solve for the position of the node, I simply divided the wavelength by two.

This strategy did not achieve the answer for either question above, and I am confused as to how this logic does not work. Any help is greatly appreciated!
 
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  • #2
You can find the frequency directly from the equations that you are given. What is the significance of the number 1300?
Kara4566 said:
Therefore, the node should be located when t=1/2(period), or 0.00245s.
This doesn't make sense. Nodes are fixed in space and time.
Kara4566 said:
To solve for the position of the node, I simply divided the wavelength by two.
This gives you the distance between adjacent nodes, not the position of the first node. If at x = 0 you have an antinode, at what fraction of the wavelength will you have the first node?
 
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Related to Pressure Waves in an Open Tube

1. What is a pressure wave in an open tube?

A pressure wave in an open tube is a propagating disturbance or change in pressure caused by the movement of a fluid, typically air, through an open tube. It can also refer to the resulting oscillation of pressure within the tube.

2. How are pressure waves generated in an open tube?

Pressure waves in an open tube are generated by a source, such as a speaker or a vibrating object, which creates fluctuations in the air pressure. These fluctuations travel through the tube and can be heard as sound waves.

3. What factors affect the speed of pressure waves in an open tube?

The speed of pressure waves in an open tube is affected by the temperature, density, and elasticity of the fluid through which they are traveling. It is also influenced by the length and diameter of the tube.

4. How do pressure waves behave in an open tube?

Pressure waves in an open tube behave as longitudinal waves, meaning that they travel in the same direction as the movement of the particles in the medium. They also experience reflection, refraction, and interference, similar to other types of waves.

5. What are some real-world applications of pressure waves in an open tube?

Pressure waves in an open tube have many practical applications, such as in musical instruments, communication systems, and medical devices. They are also used in industries such as aerospace and automotive for measuring and controlling pressure levels.

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