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fundamental frequency question, 10 pts!?
A pipe open only at one end has a fundamental frequency of 256 Hz. A 2nd pipe, initially identical to the 1st pipe, is shortened by cutting off a portion of the open end. Now, when both pipes vibrate at their fundamental frequencies, a beat frequency of 12 Hz is heard. How many cm were cut off the end of the 2nd pipe?? The speed of sound is 343 m/s
tried and tried but couldnt figure it out, help pleease :D!
1 Answer
- Steve4PhysicsLv 78 years agoFavorite Answer
v = fλ, so λ = v/f, so 256Hz corresponds to a wavelength λ = 343/256 = 1.34m
If L is the length of the first pipe L = λ/4 (see link)
L= 1.34/4 = 0.335m
The shortened pipe (length L') will have a higher fundamental frequency = 256 +12 = 268Hz
This give a new wavelength λ' = 343/268 = 1.28m
L' = 1.28/4 = 0.32m
So the length cut off = 0.335 - 0.32 = 0.015m = 1.5cm
