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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 |
| Answer» <html><body><p></p>Solution :(a)Only odd <a href="https://interviewquestions.tuteehub.com/tag/harmonics-1016029" style="font-weight:bold;" target="_blank" title="Click to know more about HARMONICS">HARMONICS</a> can exist in a pipe with one <a href="https://interviewquestions.tuteehub.com/tag/open-586655" style="font-weight:bold;" target="_blank" title="Click to know more about OPEN">OPEN</a> end. The resonant frequencies are given by f = nv/4L for harmonic numbers n = 1,<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>,5,....<br/>Because resonant frequency increases with increasing n, the second lowest resonant frequency <a href="https://interviewquestions.tuteehub.com/tag/corresponds-935737" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPONDS">CORRESPONDS</a> to n = 3, the second lowest <a href="https://interviewquestions.tuteehub.com/tag/choice-915996" style="font-weight:bold;" target="_blank" title="Click to know more about CHOICE">CHOICE</a> of n.<br/>(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,... <br/>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<br/>`f= (2v)/(2L_B)` <br/> Solving for `L_B` and substituting known data yield<br/>`L_B = v/f = (1522 m//s)/(761 Hz) = 2.00 m `</body></html> | |