A pipe that is 0.46 m long and open at both ends vibrates in the second overtone with a frequency of 1150 Hz. In this situation, the distance from the center of the pipe to the nearest antinode is closest to
A) 7.7 cm.
B) 3.8 cm.
C) 12 cm.
D) 15 cm.
E) zero.

Question

Quizplus · Accepted Answer

The distance from the center of the pipe to the nearest antinode is one-fourth of the wavelength. The wavelength is given by:$$\lambda = \frac{2L}{n+1}$$where L is the length of the pipe and n is the overtone number. In this case, L = 0.46 m and n = 2, so:$$\lambda = \frac{2(0.46 m)}{2+1} = 0.307 m$$Therefore, the distance from the center of the pipe to the nearest antinode is:$$\frac{\lambda}{4} = \frac{0.307 m}{4} = 0.077 m = 7.7 cm$$

A Pipe That Is 0