The frequency of the first overtone of a closed pipe of length l1 is e...
Overview:
The question asks us to find the ratio of the lengths of a closed pipe and an open pipe when the frequency of their first overtone is equal. In order to solve this problem, we need to understand the concept of overtones and the relationship between the length of a pipe and its resonant frequencies.
Understanding Overtones:
- Overtones are the higher harmonics produced by a vibrating object.
- The first overtone is the second harmonic, which has a frequency twice that of the fundamental frequency.
- For a closed pipe, the fundamental frequency is the frequency of the first harmonic, and the first overtone is the frequency of the second harmonic.
- For an open pipe, the fundamental frequency is the frequency of the first harmonic, and the first overtone is the frequency of the third harmonic.
Resonant Frequencies of Pipes:
- For a closed pipe, the resonant frequencies are given by the equation fn = n(v/2L), where n is the harmonic number, v is the speed of sound, and L is the length of the pipe.
- For an open pipe, the resonant frequencies are given by the equation fn = n(v/2L), where n is the harmonic number, v is the speed of sound, and L is the length of the pipe.
Solving the Problem:
- Let the length of the closed pipe be l1 and the length of the open pipe be l2.
- The frequency of the first overtone of the closed pipe is 2(v/2l1), and the frequency of the first overtone of the open pipe is 3(v/2l2).
- Since the frequencies of the first overtones are equal, we can equate the two expressions:
2(v/2l1) = 3(v/2l2)
- Canceling out common factors, we get:
l2/l1 = 2/3
- Therefore, the ratio of the lengths of the closed pipe to the open pipe is 3:2, which is equivalent to 3:4.
Answer:
The correct answer is option D) 3:4.
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