## Mastering Physics Solutions: Exercise 14.64

Exercise 14.64

Solutions Below:

A closed organ pipe has a fundamental frequency of 528 Hz (a C note) at 20° C.

Part A

What is the fundamental frequency of the pipe when the temperature is 2° C?

We know that the speed of sound, v, = λf. Start by finding the wavelength (λ) for the fundamental at 20° C:

The speed of sound is given by:

v = 331.4 * sqrt((273.15 + C) / 273.15)
v = 331.4 * sqrt((273.15 + 20) / 273.15)
v = 343m/s

Now find λ:

v = λf
λ = v / f
λ = 343 / 528
λ = 0.65

Since the wavelength of the fundamental depends on the length of the pipe, it won’t change with temperature. So it’s easy to find the new frequency.

Start by finding the speed of sound at 2° C:

v = 331.4 * sqrt((273.15 + C) / 273.15)
v = 331.4 * sqrt((273.15 + 2) / 273.15)
v = 332.6m/s

Now solve for the frequency:

v = λf
332.6 = 0.65f

f = 511.7Hz

Note: Mastering physics took an answer of 511 Hz for me. So which way you round is up to you, but if it doesn’t take your first answer, try rounding the other way:

f = 511 Hz

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