Pipe A, which is 1.80 m long and open at both ends, oscillates at its third lowe
ID: 585311 • Letter: P
Question
Pipe A, which is 1.80 m long and open at both ends, oscillates at its third lowest harmonic frequency. It is filled with air for which the speed of sound is 343 m/s. Pipe B, which is closed at one end, oscillates at its second lowest harmonic frequency. This frequency of B happens to match the frequency of A. An x axis extends along the interior of B, with x = 0 at the closed end. (a) How many nodes are along that axis? What are the (b) smallest and (c) second smallest value of x locating those nodes? (d) What is the fundamental frequency of B?
a) 2 nodes
b) 0m
Can't figure out c and d.
Explanation / Answer
(d) Fundamental frequency of the pipe
v = Velocity / 2L = 343/(2*1.8) = 95.2 Hz
(c) Now wavelength = Speed/Frequency = 3.6 m
We know that the second smallest value wil be = Wavelength / 2 = 1.8
Hence at X =1.8 m we have second node.