Pipe A , which is 1.80 m long and open at both ends, oscillates at its third low
ID: 1647484 • 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?
Explanation / Answer
length is l = 1.8 m
for open pipe third harmonic frequency is f3 = v/lamda = 3V/2L = 3*343/(2*1.8) = 285.83 Hz
for closed pipe,second lowest harmonic frequency is (3v/*4L) = 285.83 Hz
3*343/(4*L) = 285.83
L = 0.9 m is the length of closed pipe
a) for second lowest harmonic ,there are two nodes
b) smallest x value of the node is x = 0 m
c) second smallest is at x = lamda/2 = (4*L)/(3*2) = (4*0.9)/(3*2)= 0.6 m
d) fundamnetal frequency is = 285.83/3 = 95.28 Hz