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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

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

  • physics-never mind i got it!!! -

    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.

  • physics -

    wow you are smart because you figure out your own queston

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