Question

The length of the auditory canal in humans averages about 2.5 cm. What is the second lowest
standing wave frequency for a pipe of this length open at one end? Use 345 m/s for the speed of
sound.

I know that 4L=lambda, so lambda= 10cm.
I know that f=v/lambda, so f=3450 Hz.

But is that the lowest standing frequency? And how do I find the second lowest? Thanks

Answer 1

I used the eq. f=n(v/4L). Where n=3 bc that is the second lowest frequency in a tube with one end open.

Answer 2

wow you are smart because you figure out your own queston