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Pruzhinkin
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Homework Statement
Hollow tube chimes are made of metal and are open at each end. One chime is 0.54 m long.
A) if the speed of sound is 346 m/s, what is the frequency of sound produced by the third resonant length?
B)What would happen to the frequency of sound produced by the third resonant length if the chime were shorter?
Homework Equations
L1 = lambda/2 or lambda = 2L
L2 = lambda
L3 = 3 * lambda/2 or lambda = 2L/ 3
f1 = v/ lambda
The Attempt at a Solution
A) Lambda = 2L = 2 x (0.54m) = 1.08 m
f1 = v/ lambda = 346m/s / 1.08m = 320.4 Hz
L3 = 3 * (1.08m)/2 = 1.62 m
answer is
f = 346 / 1.62 = 213.6 Hz
or am i supposed to used f3=3f1, or is that only for string resonance. If it is used here then f3 = 3 * 320.4 = 961.2 Hz?
B) Not quite sure here, I understand what happens if a wavelength, frequency, or speed of sound changes, but for some reason this escapes me. I think i just fell into some sort of stupor where my head barely thinks.
Thanks in advance guys!