Homework Solution - Standing Waves and Pressure in a Closed Tube

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SUMMARY

The discussion revolves around calculating the absolute pressure at the bottom of a mercury column in a closed vertical tube, where the tube's length is 0.75 m and the air pressure is 1.01 x 10^5 Pa. The fundamental frequency of the shortened air-filled tube is equal to the third harmonic of the original tube. The final answer for the absolute pressure is determined to be 1.68 x 10^5 Pa. The equation Fn = nv/4L is referenced for understanding standing waves in tubes with one open end, but it is clarified that pressure calculations are independent of wave dynamics in this context.

PREREQUISITES
  • Understanding of standing wave equations, specifically Fn = nv/4L.
  • Knowledge of fluid mechanics, particularly hydrostatic pressure calculations.
  • Familiarity with the properties of mercury, including its mass density of 13,600 kg/m3.
  • Basic principles of acoustics related to closed and open tube configurations.
NEXT STEPS
  • Study hydrostatic pressure calculations in fluids, focusing on mercury and other liquids.
  • Explore the relationship between harmonics and standing waves in closed tubes.
  • Learn about the properties of sound waves in different mediums, including air and liquids.
  • Investigate the effects of varying tube lengths on the fundamental frequency of standing waves.
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Students in physics, particularly those studying wave mechanics and fluid dynamics, as well as educators looking for practical examples of pressure calculations in closed systems.

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Homework Statement



A vertical tube is closed at one end and open to air at the other end. The air pressure is 1.01 x 10^5 Pa. The tube has a length of 0.75 m. Mercury (mass density = 13,600 kg/m3) is poured into it to shorten the effective length for standing waves. What is the absolute pressure at the bottom of the mercury column, when the fundamental frequency of the shortened air-filled tube is equal to the third harmonic of the original tube?
ANS: 1.68 x 10^5 Pa

Homework Equations


Ive been given the equations Fn = nv/4L for tubes with one open end, although I can't seem to find any way to relate pressure and waves, can anyone help?
 
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hi 3ephemeralwnd! :smile:

(try using the X2 icon just above the Reply box :wink:)
3ephemeralwnd said:
Ive been given the equations Fn = nv/4L for tubes with one open end, although I can't seem to find any way to relate pressure and waves, can anyone help?

the pressure at the bottom of the mercury has nothing to do with the wave between the top of the mercury and the open end …

it's just a cute way of asking two totally unrelated :rolleyes: questions at once! :biggrin:

(so just use the density of mercury :wink:)
 

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