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Kara4566

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**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:

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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