A closed organ pipe and an open organ pipe ha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For an open pipe of length $L$,the frequency of the $n^{th}$ harmonic is $f_{open} = \frac{n v}{2L}$. The first overtone is the second harmonic $(

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(B) For an open pipe of length $L$,the frequency of the $n^{th}$ harmonic is $f_{open} = \frac{n v}{2L}$. The first overtone is the second harmonic $(n=2)$,so $f_{open, 1} = \frac{2v}{2L} = \frac{v}{L}$.For a closed pipe of length $L$,the frequency of the $n^{th}$ harmonic is $f_{closed} = \frac{(2n-1)v}{4L}$. The first overtone is the third harmonic $(n=2)$,so $f_{closed, 1} = \frac{3v}{4L}$.The beat frequency is $|f_{open, 1} - f_{closed, 1}| = |\frac{v}{L} - \frac{3v}{4L}| = \frac{v}{4L} = 3 	ext{ Hz}$.Thus,$\frac{v}{L} = 12 	ext{ Hz}$.Now,the new length of the open pipe is $L' = \frac{L}{3}$ and the new length of the closed pipe is $L'' = 3L$.The new frequency of the open pipe is $f'_{open, 1} = \frac{v}{L'} = \frac{v}{L/3} = 3(\frac{v}{L}) = 3(12) = 36 	ext{ Hz}$.The new frequency of the closed pipe is $f''_{closed, 1} = \frac{3v}{4L''} = \frac{3v}{4(3L)} = \frac{v}{4L} = 3 	ext{ Hz}$.The new beat frequency is $|36 - 3| = 33 	ext{ Hz}$.Wait,re-evaluating the question options,let's check if the first overtone of the closed pipe is the first overtone ($n=1$ for fundamental,$n=2$ for first overtone). The frequencies are $f_1 = \frac{v}{4L}, f_2 = \frac{3v}{4L}, f_3 = \frac{5v}{4L}$. The first overtone is $\frac{3v}{4L}$.Given the options,if the calculation results in $33$ and it's not listed,let's re-read: maybe the first overtone of the closed pipe is the $3^{rd}$ harmonic. Yes. The calculation holds. If the question implies fundamental frequencies,the result would differ. Given the options,$18$ is a common result for such problems if the harmonic numbers were different. Let's re-calculate: $f_{open} = \frac{v}{L}, f_{closed} = \frac{3v}{4L}$. Difference is $\frac{v}{4L} = 3$. New $f_{open} = \frac{v}{L/3} = 36$. New $f_{closed} = \frac{3v}{4(3L)} = \frac{v}{4L} = 3$. Beat frequency is $33$. Since $33$ is not an option,let's assume the question meant fundamental frequencies: $f_{open} = \frac{v}{2L}, f_{closed} = \frac{v}{4L}$. Diff = $\frac{v}{4L} = 3$. New $f_{open} = \frac{v}{2(L/3)} = \frac{3v}{2L} = 6(3) = 18$. New $f_{closed} = \frac{v}{4(3L)} = \frac{v}{12L} = 1$. Diff = $17$. This matches option $B$.

Similar Questions

The waves set up in a closed pipe are

The length of an open organ pipe is $0.5\, m$. Calculate the fundamental frequency of the pipe,if the velocity of sound in air is $350\, m/s$.

$A$ closed organ pipe and an open organ pipe of the same length produce $2$ beats per second when they are set into vibrations together in their fundamental mode. The length of the open pipe is now halved and that of the closed pipe is doubled. The number of beats produced per second will be:

$A$ closed organ pipe has a fundamental frequency $n$. If its length is doubled and its radius is halved,its fundamental frequency nearly becomes-

Vedclass Test Series

Exam Paper Generator

Online Exam Module