Question

explain please

*Question 160: Resonant Frequencies A piece of pipe is closed at one end and open at the other. The standing wave with the lowest frequency (the fundamental) occurs at frequency 100 Hz. What is the frequency of the first overtone (the next highest standing wave frequency)? Select one: a. 200 Hz b. 150 Hz c. No other standing waves are possible in a pipe open at one end only d. 67 Hz e. 300 Hz The correct answer is: 300 Hz

Answer 1

Fia an argon pipe at close end there will always node and at open always antinode.so for fundamental frequency there will be like shown in figure 1 . So it is clearly that the length of pipe is equal to the 1/4 of the wavelength as calculated and for first overtone the pipe length will be the same but, wave will like shown in figure 2 and so clearly the length of th pipe will 3/4 of the wavelength in this scenario. And it is also calculated above. and so by the relation ( velocity=frequency×wavelength ) relate both frequencies because the velocity is same in the Both case.

امن و Pipe is closed at one end and open at another So at Close end it has node always and at open end it will autined always So. for fundamental frequency it will be like first first case and second case (first overtone) < it will be like second figure. so from Ist fiqur. e=1 ) x1 = 40 tiy O so, frequency, U = f (CV) and for first ouen tone. l = 311, daud after the go. U, a ea, D devided on by cao = 1 1 2 3 u 1 = 4 2 = 3u1 & 2 2 34 100 CEV. 2000Ha given) 2 2 800 by un