A tube with a cap on one end, but open at the other end, has a fundamental frequency of 129.5 Hz. The speed of sound is 343 m/s. (a) If the cap is removed, what is the new fundamental frequency of the tube? (b) How long is the tube?

(a) 391 Hz.

(b) 0.439 m

Explanation:

(a)

The fundamental frequency of a closed pipe is given as

fc = v/4l ... Equation 1

Where fc = fundamental frequency, v = velocity of sound, l = length of pipe

make l the subject of the equation

l = v/4fc ... Equation 2

Also for an open pipe,

fo = v/2l ... Equation 3

Where fo = fundamental frequency of open pipe.

making l the subject of the equation.

l = v/2fo ... Equation 4

Therefore, equating equation 4 and equation 2

v/2fo = v/4fc

fo = 2fc ... Equation 5.

Note: When the cap is removed, the tube becomes an open tube.

Given: fc = 195.5 Hz.

Substitute into equation 5

fo = 2 (195.5)

fo = 391 Hz.

(b)

using equation 2 above,

l = v/4fc

Given: v = 343 m/s, fc = 195.5 Hz.

l = 343 / (4*195.5)

l = 343/782

l = 0.439 m.

Hence the tube is 0.439 m long.