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) The frequency of the $m^{th}$ overtone for a closed organ pipe is given by $f_{c} = \frac{(2m+1)V}{4L_{c}}$,where $m=1$ for the first overtone,so

Vedclass · Accepted Answer

$3:4$

Vedclass · Answer

(B) The frequency of the $m^{th}$ overtone for a closed organ pipe is given by $f_{c} = \frac{(2m+1)V}{4L_{c}}$,where $m=1$ for the first overtone,so $f_{c} = \frac{3V}{4L_{c}}$.The frequency of the $m^{th}$ overtone for an open organ pipe is given by $f_{o} = \frac{(m+1)V}{2L_{o}}$,where $m=1$ for the first overtone,so $f_{o} = \frac{2V}{2L_{o}} = \frac{V}{L_{o}}$.Given that the frequencies are identical,$f_{c} = f_{o}$:$\frac{3V}{4L_{c}} = \frac{V}{L_{o}}$Rearranging the terms to find the ratio of lengths $L_{c}:L_{o}$:$\frac{L_{c}}{L_{o}} = \frac{3}{4}$Thus,the ratio of their lengths is $3:4$.

$A$ closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio:

Similar Questions

The frequency of the first overtone of a closed pipe of length $l_{1}$ is equal to that of the first overtone of an open pipe of length $l_{2}$. The ratio of their lengths $(l_{1}: l_{2})$ is

Discuss stationary waves in pipes and explain the types of pipes.

If the velocity of sound in air is $336 \ m/s$,the maximum length of a closed pipe that would produce a just audible sound will be:

$A$ cylindrical tube $(L = 120 \, cm)$ is resonant with a tuning fork of frequency $330 \, Hz$. If it is being filled with water,then to get resonance again,the minimum length of the water column is ...... $cm$ $(v_{air} = 330 \, m/s)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module