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An organ pipe, closed at one and open at the other, is designed to have a fundamental frequency of c( 131 Hz) what is the frequency of the next higher harmonic for this pipe?
Correct answer is '393'. Can you explain this answer?
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An organ pipe, closed at one and open at the other, is designed to hav...
For a pipe of length L, Closed at one and open at the other end, the resonant frequencies are given by
where n - 1, 3, 5
and v is the speed of the sound
Thus, a fundamental frequency is
and the next harmonic has a frequency 
For the next harmonic,
f(n = 3) = 3f(n = 1) = 3 x 131
= 393 Hz
The correct answer is: 393
where n - 1, 3, 5and v is the speed of the sound
Thus, a fundamental frequency is
and the next harmonic has a frequency For the next harmonic,f(n = 3) = 3f(n = 1) = 3 x 131
= 393 Hz
The correct answer is: 393
Most Upvoted Answer
An organ pipe, closed at one and open at the other, is designed to hav...
Frequency of the Fundamental Mode
The fundamental frequency of an organ pipe closed at one end and open at the other is given by the equation:
f = (v/4L)
where f is the frequency, v is the speed of sound, and L is the length of the pipe.
Given that the fundamental frequency is 131 Hz, we can rearrange the equation to solve for the speed of sound (v):
v = 4Lf
Substituting the given values, we have:
v = 4 * L * 131 Hz
Frequency of the Next Higher Harmonic
The frequency of the next higher harmonic can be found by multiplying the fundamental frequency by the harmonic number. In this case, we are looking for the second harmonic:
f2 = 2 * f1
where f2 is the frequency of the second harmonic, and f1 is the frequency of the fundamental mode.
Substituting the given value for the fundamental frequency, we have:
f2 = 2 * 131 Hz
Calculation
To find the frequency of the next higher harmonic, we substitute the value of the fundamental frequency into the equation for the second harmonic:
f2 = 2 * 131 Hz = 262 Hz
However, this is not the correct answer given in the question. Therefore, we need to re-evaluate the given information.
Correction
The correct frequency of the next higher harmonic for this pipe is given by the equation:
f2 = 3 * f1
where f2 is the frequency of the second harmonic, and f1 is the frequency of the fundamental mode.
Substituting the given value for the fundamental frequency, we have:
f2 = 3 * 131 Hz = 393 Hz
Therefore, the frequency of the next higher harmonic for this pipe is 393 Hz.
The fundamental frequency of an organ pipe closed at one end and open at the other is given by the equation:
f = (v/4L)
where f is the frequency, v is the speed of sound, and L is the length of the pipe.
Given that the fundamental frequency is 131 Hz, we can rearrange the equation to solve for the speed of sound (v):
v = 4Lf
Substituting the given values, we have:
v = 4 * L * 131 Hz
Frequency of the Next Higher Harmonic
The frequency of the next higher harmonic can be found by multiplying the fundamental frequency by the harmonic number. In this case, we are looking for the second harmonic:
f2 = 2 * f1
where f2 is the frequency of the second harmonic, and f1 is the frequency of the fundamental mode.
Substituting the given value for the fundamental frequency, we have:
f2 = 2 * 131 Hz
Calculation
To find the frequency of the next higher harmonic, we substitute the value of the fundamental frequency into the equation for the second harmonic:
f2 = 2 * 131 Hz = 262 Hz
However, this is not the correct answer given in the question. Therefore, we need to re-evaluate the given information.
Correction
The correct frequency of the next higher harmonic for this pipe is given by the equation:
f2 = 3 * f1
where f2 is the frequency of the second harmonic, and f1 is the frequency of the fundamental mode.
Substituting the given value for the fundamental frequency, we have:
f2 = 3 * 131 Hz = 393 Hz
Therefore, the frequency of the next higher harmonic for this pipe is 393 Hz.
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An organ pipe, closed at one and open at the other, is designed to have a fundamental frequency of c( 131 Hz) what is the frequency of the next higher harmonic for this pipe?Correct answer is '393'. Can you explain this answer?
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An organ pipe, closed at one and open at the other, is designed to have a fundamental frequency of c( 131 Hz) what is the frequency of the next higher harmonic for this pipe?Correct answer is '393'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about An organ pipe, closed at one and open at the other, is designed to have a fundamental frequency of c( 131 Hz) what is the frequency of the next higher harmonic for this pipe?Correct answer is '393'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An organ pipe, closed at one and open at the other, is designed to have a fundamental frequency of c( 131 Hz) what is the frequency of the next higher harmonic for this pipe?Correct answer is '393'. Can you explain this answer?.
An organ pipe, closed at one and open at the other, is designed to have a fundamental frequency of c( 131 Hz) what is the frequency of the next higher harmonic for this pipe?Correct answer is '393'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about An organ pipe, closed at one and open at the other, is designed to have a fundamental frequency of c( 131 Hz) what is the frequency of the next higher harmonic for this pipe?Correct answer is '393'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An organ pipe, closed at one and open at the other, is designed to have a fundamental frequency of c( 131 Hz) what is the frequency of the next higher harmonic for this pipe?Correct answer is '393'. Can you explain this answer?.
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