Psych 3A03
McMaster University
28 September 2009
Paul A. Faure
* Problem Set 3 - Additional Information for Question 10 *
Resonance In a Tube Open At Both Ends or Closed at Both Ends
As explained in part three of lecture 3, when a tube is closed at both ends or
open at both ends the wavelength of the standing wave of the lowest resonant
frequency (i.e. the fundamental frequency (f0) of vibration) in the tube is equal to
twice the length (L) of the tube (= 2L). Higher frequencies at integer multiples
of the fundamental frequency are also able to form standing waves in a tube that
is open or closed at both ends. A standing wave in a tube that is open at both
ends must have vibrational anti-nodes at both ends. A standing wave in a tube
closed at both ends must have vibrational nodes at both ends. Thus, the
fundamental and first two harmonic resonant frequencies of a tube that is open or
closed at both ends are:
f0 = c/ ; f0 = c/(2L) (fundamental frequency)
f2 = 2f0 (second harmonic)
f3 = 3f0 (third harmonic)
Resonance In a Tube Open At One End and Closed at The Other End
When a tube is closed at one end and open at the other end (e.g. ear canal, beer
bottle), the wavelength of the lowest resonant frequency (i.e. fundamental
frequency or f0), also known as the fundamental mode of vibration, occurs when
the standing wave has a vibrational node at the closed end and an anti-node at
the open end. The lowest frequency standing wave that satisfies the condition of
having a node at the closed end and an anti-node at the open end will have a
wavelength that is four times the length (L) of the tube (= 4L). Thus, for a tube
of length L, the fundamental frequency of vibration is calculated as f0 = c/(4L).
Higher frequencies of the fundamental frequency will also be able to form
standing waves; however, this will be true only at odd integer multiples of f0
(i.e. at odd integer multiples of the fundamental there will be a vibrational node at
the closed end and an anti-node at the open end). Thus, the first three resonant
frequencies of a tube open at one end and closed at the other end will be:
f0 = c/ ; f0 = c/(4L) (fundamental frequency)
f3 = 3f0 (third harmonic, which is the first odd integer multiple of f0)
f5 = 5f0 (fifth harmonic, which is the second odd integer multiple of f0)
Problem Set #3 (Addendum)