A string of length 0.8 m has a wave speed of 320 m/s. What are the frequencies of the first three harmonics?
Think about your answer, then reveal below.
Model answer: f₁ = v/(2L) = 320/(2·0.8) = 200 Hz. f₂ = 2f₁ = 400 Hz. f₃ = 3f₁ = 600 Hz. All integer harmonics are present because the string is fixed at both ends.
The fundamental formula for a string (or open pipe) is f₁ = v/(2L). Each higher harmonic is an integer multiple: fₙ = nf₁. For a fixed-fixed string, n = 1, 2, 3, ... with no missing harmonics.
Question 2 Short Answer
Why does a pipe closed at one end only support odd harmonics?
Think about your answer, then reveal below.
Model answer: A closed end requires a displacement node; an open end requires an antinode. The only wavelengths that satisfy node-at-one-end and antinode-at-the-other end are those where the pipe length equals an odd multiple of a quarter-wavelength: L = λ/4, 3λ/4, 5λ/4... This gives frequencies fₙ = nf₁ where n = 1, 3, 5 only. Even multiples would require the same boundary condition at both ends, which isn't the case here.
Contrast with an open pipe (antinode at both ends) or a fixed string (node at both ends) — those symmetric conditions allow both even and odd harmonics. The asymmetric boundary conditions of the closed pipe are what eliminate the even harmonics.
Question 3 Multiple Choice
An open organ pipe has a fundamental frequency of 120 Hz. A closed pipe of the same length is played. What is its fundamental frequency, and what harmonics does it produce?
A120 Hz; all integer harmonics
B60 Hz; odd harmonics only
C240 Hz; even harmonics only
D60 Hz; all integer harmonics
A closed pipe's fundamental is v/(4L), while an open pipe's is v/(2L). Same length means the closed fundamental is exactly half: 60 Hz. And because of the asymmetric boundary conditions, closed pipes only produce odd harmonics: 60 Hz, 180 Hz, 300 Hz, ...