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Two identical tubes, each closed at one end, have a fundamental frequency of 305 Hz at...

Two identical tubes, each closed at one end, have a fundamental frequency of 305 Hz at 25.0 ∘C. The air temperature is increased to 30.0 ∘C in one tube.

If the two pipes are now sounded together, what beat frequency results?

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Answer #1

T1 = 273 + 25 = 298 K

T2 = 273 + 30 = 303 K

f1 = 305 Hz

f = v/lambda

v= v0*sqrt T

So,

f is proportional to sqrt T

f1/f2 = sqrt (T1/T2)

f2 = f1*sqrt (T2/T1)

f2 = 305*sqrt (303/298) = 307.55 Hz

Beat frequency = 307.55 - 305 = 2.55 Hz

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