Homework Solution - Standing Waves and Pressure in a Closed Tube

[3ephemeralwnd]

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?

 

[tiny-tim]

hi 3ephemeralwnd!

(try using the X² icon just above the Reply box )

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 questions at once!

(so just use the density of mercury )