A closed organ pipe of radius r_1 and an open | vedclass

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

Question

(C) For a closed organ pipe with end correction,the frequency of the $n^{th}$ mode is given by $
u_{n}^{\prime} = \frac{(2n-1)v}{4(L+0.6r_1)}$,where

Vedclass · Accepted Answer

$r_2 - 2r_1 = 2.5L$

Vedclass · Answer

(C) For a closed organ pipe with end correction,the frequency of the $n^{th}$ mode is given by $
u_{n}^{\prime} = \frac{(2n-1)v}{4(L+0.6r_1)}$,where $0.6r_1$ is the end correction.For the fundamental mode $(n=1)$,$
u_{1}^{\prime} = \frac{v}{4(L+0.6r_1)}$.For an open organ pipe with end correction,the frequency of the $m^{th}$ mode is given by $
u_{m} = \frac{mv}{2(L+2 	imes 0.6r_2)}$,where $0.6r_2$ is the end correction at each end.For the first overtone $(m=2)$,$
u_{2} = \frac{2v}{2(L+1.2r_2)} = \frac{v}{L+1.2r_2}$.Since both pipes resonate with the same tuning fork,$
u_{1}^{\prime} = 
u_{2}$.Therefore,$\frac{v}{4(L+0.6r_1)} = \frac{v}{L+1.2r_2}$.$L + 1.2r_2 = 4L + 2.4r_1$.$1.2r_2 - 2.4r_1 = 3L$.Dividing by $1.2$,we get $r_2 - 2r_1 = 2.5L$.

$A$ closed organ pipe of radius $r_1$ and an open organ pipe of radius $r_2$ having the same length $L$ resonate when excited with a given tuning fork. The closed organ pipe resonates in its fundamental mode,whereas the open organ pipe resonates in its first overtone. Then:

Similar Questions

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is $60 \,cm$, the length of the closed pipe will be: (in $\,cm$)

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 $350\, m/s$.

An open air pipe of length $80 \,cm$ has the second harmonic frequency equal to the fundamental frequency of a closed organ air pipe. The length of the closed pipe is (in $\,cm$)

An open tube is in resonance with a string (frequency of vibration of the tube is $n_0$). If the tube is dipped in water so that $75\%$ of the length of the tube is inside the water,then the ratio of the frequency of the tube to the string now will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module