Standing waves and resonance in a glass tube

[kapitan90]

Homework Statement

A glass tube is closed at one end. The air column it contains has a length that can be varied
between 0.50m and 1.50 m. If a tuning fork of frequency 306 Hz is sounded at the top of
the tube, at which lengths of the air column would resonance occur? (Take the speed of
sound to be 330 ms-1.)

Homework Equations

When we want to check where resonance occurs, how do we know whether the standing wave created is first, second or third harmonic?

 

[jbsteele]

The Attempt at a SolutionIn this case, the fundamental frequency (f1) of the standing wave is equal to the frequency of the tuning fork, which is 306 Hz. The first harmonic (f2) is twice the fundamental frequency (2f1=612Hz), the second harmonic (f3) is three times the fundamental frequency (3f1=918Hz) and the third harmonic (f4) is four times the fundamental frequency (4f1=1224Hz). To find the length of the air column at which resonance occurs, we can use the equation: L= (n*v)/f where L is the length of the air column, n is the harmonic number (1,2,3,4, etc.), v is the speed of sound (330 ms-1) and f is the frequency of the standing wave. Therefore, for the first harmonic, the length of the air column at which resonance occurs is:L1= (1*330)/306 = 1.084 m For the second harmonic, the length of the air column at which resonance occurs is:L2= (2*330)/306 = 2.168 m For the third harmonic, the length of the air column at which resonance occurs is:L3= (3*330)/306 = 3.252 m Therefore, resonance will occur when the length of the air column is 1.084 m, 2.168 m or 3.252 m.