Question 421: Waves

Question

The fundamental frequency of an air column in pipe 'A' closed at one end coincides with second overtone of pipe 'B' open at both ends. The ratio of length of pipe 'A' to that of pipe 'B' is

Accepted Answer

To solve this problem, let's first understand the fundamental frequencies and harmonics in the two types of pipes described.
Pipe 'A' is closed at one end. For such a pipe, the fundamental frequency (first harmonic) is given by:
$$ f_A = \frac{v}{4L_A} $$
where $ v $ is the speed of sound and $ L_A $ is the length of the pipe. The harmonics in a closed pipe are odd multiples of the fundamental frequency. So, the frequencies are:
$$ f_A, 3f_A, 5f_A, \ldots $$
Pipe 'B' is open at both ends. For such a pipe, the fundamental frequency (first harmonic) is given by:
$$ f_B = \frac{v}{2L_B} $$
where $ L_B $ is the length of the pipe. The harmonics in an open pipe are whole number multiples of the fundamental frequency. So, the frequencies are:
$$ f_B, 2f_B, 3f_B, \ldots $$
According to the problem, the fundamental frequency of pipe 'A' coincides with the second overtone of pipe 'B'. The second overtone of pipe 'B' is the third harmonic, which is:
$$ 3f_B = 3 \left(\frac{v}{2L_B}\right) = \frac{3v}{2L_B} $$
We set this equal to the fundamental frequency of pipe 'A':
$$ \frac{v}{4L_A} = \frac{3v}{2L_B} $$
By canceling the common terms and rearranging, we get:
$$ \frac{1}{4L_A} = \frac{3}{2L_B} $$
Cross-multiplying gives:
$$ 2L_B = 12L_A $$
Simplifying, we find:
$$ L_B = 6L_A $$
Thus, the ratio of the length of pipe 'A' to that of pipe 'B' is:
$$ \frac{L_A}{L_B} = \frac{L_A}{6L_A} = \frac{1}{6} $$
Therefore, the ratio is:
Option C 1 : 6

Answer

3 : 8

Answer

3 : 4

Answer

1 : 6

Answer

2 : 3

Waves

The fundamental frequency of an air column in pipe 'A' closed at one end coincides with second overtone of pipe 'B' open at both ends. The ratio of length of pipe 'A' to that of pipe 'B' is

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