An open pipe,0.95 m long,vibrates in the second overtone with a frequency of 561 Hz.In this situation,the distance from the center of the pipe to the nearest antinode,in cm,is closest to:
A)16
B)7.9
C)24
D)32
E)zero

Question

Quizplus · Accepted Answer

The second overtone has 3 antinodes and 2 nodes. The distance between adjacent antinodes is 1/4 of the wavelength. The wavelength is twice the length of the pipe, so the distance between adjacent antinodes is 1/2 of the length of the pipe. Therefore, the distance from the center of the pipe to the nearest antinode is 1/4 of the length of the pipe, which is 0.95 m / 4 = 0.2375 m = 23.75 cm. The closest choice is A) 16 cm.

An Open Pipe,0