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Chapter 16: Problem 47

A standing wave in a pipe with both ends open has a frequency of \(440 \mathrm{~Hz}\). The next higher harmonic has a frequency of \(660 \mathrm{~Hz}\) a) Determine the fundamental frequency. b) How long is the pipe?

Short Answer

Expert verified
Question: Given the frequencies of a standing wave and its next higher harmonic in an open pipe to be 440 Hz and 660 Hz, determine the pipe's fundamental frequency and length. Answer: The fundamental frequency is 220 Hz, and the length of the pipe is approximately 0.780 meters.

Step by step solution

01

Understanding the relationship between harmonics and fundamental frequency

In an open pipe, the frequencies of the harmonics are integer multiples of the fundamental frequency. For the given frequencies, we can express them in terms of the fundamental frequency: \(440Hz = n_1 \times f_1\) \(660Hz = n_2 \times f_1\) Since \(n_1\) and \(n_2\) are integers, we should find two integers that fit the equation.
02

Determine the fundamental frequency

Start by finding the difference between the two given frequencies: \(\Delta f = 660 Hz - 440 Hz = 220 Hz\) Now, let's find their ratio: \(\frac{660Hz}{440 Hz} = \frac{3}{2}\) The ratio of 3 to 2 means that we are given 2nd and 3rd harmonics. Now we can write the equations for the 2nd and 3rd harmonics: \(440Hz = 2 \times f_1\) \(660Hz = 3 \times f_1\) To find the fundamental frequency, we will divide the frequency of 2nd harmonic by the integer (n=2) according to the equation: \(F_1 = \frac{440Hz}{2} = 220 Hz\) Thus, the fundamental frequency is \(220 Hz\).
03

Determine the relationship of length and frequency for open pipe

To find the length of the pipe, we should use the equation for the fundamental frequency of open pipes: \(f_1 = \frac{c}{2L}\) Where: - \(f_1\) is the fundamental frequency - \(c\) is the speed of sound in the pipe (assuming room temperature, \(c \approx 343 m/s\)) - \(L\) is the length of the pipe
04

Calculate the length of the pipe

Now that we have found the fundamental frequency, we can rearrange the formula, solving for L: \(L = \frac{c}{2f_1}\) Substitute the known values: \(L = \frac{343}{2 \times 220}\) \(L \approx 0.780 \mathrm{~m}\) So the length of the open pipe is approximately \(0.780\) meters.

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Most popular questions from this chapter

A half-open pipe is constructed to produce a fundamental frequency of \(262 \mathrm{~Hz}\) when the air temperature is \(22^{\circ} \mathrm{C} .\) It is used in an overheated building when the temperature is \(35^{\circ} \mathrm{C} .\) Neglecting thermal expansion in the pipe, what frequency will be heard?

Two identical half-open pipes each have a fundamental frequency of \(500 .\) Hz. What percentage change in the length of one of the pipes will cause a beat frequency of \(10.0 \mathrm{~Hz}\) when they are sounded simultaneously?

A sound meter placed \(3 \mathrm{~m}\) from a speaker registers a sound level of \(80 \mathrm{~dB}\). If the volume on the speaker is then turned down so that the power is reduced by a factor of 25 , what will the sound meter read? a) \(3.2 \mathrm{~dB}\) c) \(32 \mathrm{~dB}\) e) \(66 \mathrm{~dB}\) b) \(11 \mathrm{~dB}\) d) \(55 \mathrm{~dB}\)

If you blow air across the mouth of an empty soda bottle, you hear a tone. Why is it that if you put some water in the bottle, the pitch of the tone increases?

A (somewhat risky) way of telling if a train that cannot be seen or heard is approaching is by placing your ear on the rail. Explain why this works.
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