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
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?