   Chapter 14, Problem 66P

Chapter
Section
Textbook Problem

# Two pipes of equal length are each open at one end. Each has a fundamental frequency of 480. Hz at 300. K. In one pipe the air temperature is increased to 305 K. If the two pipes are sounded together, what beat frequency results?

To determine
The beat frequency of the sound in two pipes.

Explanation

Given Info: The temperature of the first pipe is T1=300K .

Formula to calculate the speed of the sound in both the pipes is,

v=(331.3m/s)T273K

• v is the speed of the sound in two pipes.
• T is the temperature of the two pipes.

Formula to calculate the speed of the sound in the first pipe is,

v1=(331.3m/s)T1273K

• v1 is the speed of the sound in the first pipe.
• T1 is the temperature of the first pipe.

Substitute 300K for T1 to find v1 .

v1=(331.3m/s)300K273K=347.3m/s

Formula to calculate the speed of the sound in the second pipe is,

v2=(331.3m/s)T2273K

• v2 is the speed of the sound in the second pipe.
• T2 is the temperature of the second pipe.

Substitute 305K for T2 to find v2 .

v2=(331.3m/s)305K273K=350.2m/s

Since from the given information the two pipes are having same length, so that the formula for fundamental wavelength will be,

λ=4L (1)

• λ is the fundamental wavelength.
• L is the length of the two pipes.

The relation between the speed and wavelength is,

v=fλ (2)

• f is the frequency

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