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A pipe closed at one end has a fundamental frequency of 100 Hz. What is the frequency of the next higher harmonic of the pipe?

a. 300 Hz
b. 400 Hz
c. 33 Hz
d. 200 Hz

User Loufs
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1 Answer

3 votes

Final answer:

In this case, the frequency of the next higher harmonic is 165.5 Hz. To find the frequency of the next higher harmonic of a pipe closed at one end, you can use the formula f = nv/2L, where n is the harmonic number, v is the speed of sound, and L is the length of the pipe. By calculating the length of the pipe, we can determine the frequency of the next higher harmonic.

Step-by-step explanation:

An open-pipe resonator has a fundamental frequency of 250 Hz. By changing its length, we can find the frequency of the next higher harmonic. The fundamental frequency is given by the formula f = nv/2L, where f is the frequency, n is the harmonic number (1 for fundamental), v is the speed of sound, and L is the length of the resonator. Solving for L, we can then substitute in the values for the next harmonic and find the new length. Rearranging the formula, we have L = nv/2f. Plugging in the values for n=2, v=331 m/s, and f=250 Hz, we can calculate:

  1. L = 2 * 331/2 * 250
  2. L = 663/500
  3. L = 1.326 m

Hence, the frequency of the next higher harmonic of the pipe is obtained by using the formula f = nv/2L. Since the length of the pipe is now 1.326 m, we can calculate the frequency of the next higher harmonic (n=3) as:

  1. f = 3 * 331/2 * 1.326
  2. f = 165.5 Hz

Therefore, the correct answer is option c. 165.5 Hz.

User PatrickSteele
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