Physics, asked by Rohilahuja123, 1 year ago

An open pipe is suddenly closed at one end, as a result of which the frequency of third harmonic of the closed pipe is found to be 100Hz higher than the fundamental frequency of the open pipe. If fundamental frequency of the open pipe is 400/n then the value of n is ?

Answers

Answered by christinamariyamjoy2
0

Answer: n = 8

Explanation:  Fundamental frequency of open pipe = v/2L = 400/n                       Third harmonic frequency of open pipe = 3v/4L                                                                                          Given that 3rd harmonic frequency of closed is 100 Hz higher than fundamental frequency of open.                                                                                         Therefore, 3v/4L = v/2L +100                                                                                                   v/4L = 100 i.e, v/ 2L = 50                                                                                                        400/n =50 i.e, n = 400/50 = 8

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