An open organ pipe is closed such that the third overtone of the closed pipe is found to be higher in frequency by 200 Hz than the second overtone of the original pipe. The fundamental frequency of the open pipe is (Neglect end correction)

Step-by-step solution

Let L be the length of the organ pipe and v the speed of sound in air. For an open organ pipe, the fundamental frequency is f_o = v 2L , with overtones at integer multiples. The second overtone is 3f_o . When closed, the pipe behaves as a closed organ pipe with fundamental frequency f_c = v 4L , producing odd harmonics. The third overtone is 7f_c . Given that the closed pipe's third overtone exceeds the open pipe's second overtone by 200 Hz: 7f_c = 3f_o + 200 Since f_c = 1 2 f_o , substitution yields: 7 2 f_o = 3f_