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

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.

wow you are smart because you figure out your own queston

To find the second lowest standing wave frequency for a pipe of length 2.5 cm open at one end, you can use the formula f = n*v/2L, where n is the harmonic number, v is the speed of sound, and L is the length of the pipe.

In this case, since the pipe is open at one end, the harmonics that can be produced are odd multiples of the fundamental frequency. Therefore, the first harmonic is the fundamental frequency, the second harmonic is the first overtone, the third harmonic is the second overtone, and so on.

To find the lowest standing wave frequency (or the fundamental frequency), you can substitute n = 1 into the formula:
f = (1 * 345 m/s) / (2 * 2.5 cm)

Simplifying this expression, you get:
f = 345 m/s / 5 cm
f = 69 Hz

So, the lowest standing wave frequency is 69 Hz.

To find the second lowest, or the first overtone, you substitute n = 3 into the formula:
f = (3 * 345 m/s) / (2 * 2.5 cm)

Simplifying this expression, you get:
f = 1035 m/s / 5 cm
f = 207 Hz

Therefore, the second lowest standing wave frequency for a pipe of length 2.5 cm open at one end is 207 Hz.