Question

help! dont really know how to approach this.

A pipe of length L that is open at both ends is resonating at its fundamental frequency. Which statement is true? The wavelength is L and there is a displacement node at the pipe's midpoint. The wavelength is 2L and there is a displacement node at the pipe's midpoint. The wavelength is L and there is a displacement antinode at the pipe's midpoint. The wavelength is 2L and there is a displacement antinode at the pipe's midpoint. The wavelength is 4L and there is a displacement antinode at the pipe's midpoint.

Answer 1

If a pipe is open at both ends, it is called an open pipe. In an open pipe, the frequency of fundamental note or first harmonic formed is shown below.

不 IN E * v

If v is the velocity of sound in air and $\lambda$ is the wavelength, then the fundamental frequency $\nu$ is given by

$\nu=\frac{v}{\lambda}$

If L is the length of the tube, then from the figure,

$L = \frac{\lambda}{4}+\frac{\lambda}{4}=\frac{\lambda}{2}$

$\lambda = 2L$

So the fundamental frequency of the open pipe is,

$\nu=\frac{v}{2L}$

Also from the figure, a node is formed at the midpoint.

So for an open pipe resonating at its fudamental frequency, the wavelength is 2L and there is a displacement node at the pipe's midpoint.

Second option is the correct answer.