Part A If the pipe is open at both ends, determine the locations along the pipe

Question

Part A

If the pipe is open at both ends, determine the locations along the pipe (measured from the left end) of the displacement nodes for the fundamental frequency. If more than one value, separate with commas.

x=_____m

Part B

If the pipe is open at both ends, determine the locations along the pipe (measured from the left end) of the displacement nodes for the first overtone.
If there is more than one value, separate your answers with commas.
x =_____m


Part C
If the pipe is open at both ends, determine the locations along the pipe (measured from the left end) of the displacement nodes for the second overtone.
If there is more than one value, separate your answers with commas.
x =______m


Part D -
If the pipe is closed at the left end and open at the right end , determine the locations along the pipe (measured from the left end) of the displacement nodes for the fundamental frequency.
If there is more than one value, separate your answers with commas.
x =______m


Part E
If the pipe is closed at the left end and open at the right end , determine the locations along the pipe (measured from the left end) of the displacement nodes for the first overtone.
If there is more than one value, separate your answers with commas.
x =______m


Part F
If the pipe is closed at the left end and open at the right end , determine the locations along the pipe (measured from the left end) of the displacement nodes for the second overtone.
If there is more than one value, separate your answers with commas .
x =______m

Explanation / Answer

For open pipes antinodes are formed at ends of pipe and nodes are formed inbetween two ends. (A)For fundamental frequency, dispacement node at center of pipe.      So x = L/2 = 1.10 m (B) For first overtone, dispacement node are          x = L/4 = 0.55 m           x = L - L/4             = 1.65 m Then x = 0.55m,1.65m C) For second overtone, dispacement node are           x = L/6 = 0.366 m           x = L/3 + L/6            = 0.733+ 0.366 m             = 1.099 m           x = L/3 + L/6+L/6            = 0.733+ 0.366 m +0.366m             = 1.465 m           x = L/3 + L/6+L/6            = 0.733+ 0.366 m +0.366m             = 1.465 m Then x = 0.366m,1.099m , 1.465m (D)For fundamental frequency, dispacement node at left end of pipe.      So x = 0 (B) For first overtone, dispacement node are          x = 0           x = L/3             = 0.733m Then x = 0, 0.733m C) For second overtone, dispacement node are           x = 0           x =2L/5             = 0.88           x = 2L/5+ 2L/5            = 0.733+ 0.366 m +0.366m             = 1.76 m Then x = 0,0.88m,1.76m      So x = 0 (B) For first overtone, dispacement node are          x = 0           x = L/3             = 0.733m Then x = 0, 0.733m C) For second overtone, dispacement node are           x = 0           x =2L/5             = 0.88           x = 2L/5+ 2L/5            = 0.733+ 0.366 m +0.366m             = 1.76 m           x = 2L/5+ 2L/5            = 0.733+ 0.366 m +0.366m             = 1.76 m Then x = 0,0.88m,1.76m           

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