InterviewSolution
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InterviewSolution
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The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length L_A=1.5m and one open end, contains a small sound source that sets up the standing wave with the second lowest resonant frequency of that pipe. Sound from pipe A sets up resonance in pipe B, which has both ends open. The resonance is at the second lowest resonant frequency of pipe B. The length of the pipe B is |
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Answer» Solution :(a)Only odd HARMONICS can exist in a pipe with one OPEN end. The resonant frequencies are given by f = nv/4L for harmonic numbers n = 1,3,5,.... Because resonant frequency increases with increasing n, the second lowest resonant frequency CORRESPONDS to n = 3, the second lowest CHOICE of n. (b) Any harmonic can exist in a pipe with two open ends. The resonant frequencies are given by f = n/2L for harmonic numbers n = 1,2,3,... Because resonant frequency increases with increasing n, the second lowest resonant frequency corresponds to n = 2, the second lowest choice of n. Therefore, we can write `f= (2v)/(2L_B)` Solving for `L_B` and substituting known data yield `L_B = v/f = (1522 m//s)/(761 Hz) = 2.00 m ` |
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