The second overtone of an open organ pipe A a | vedclass

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

Question

(D) For an open organ pipe $A$,the frequency of the $n$-th overtone is given by $
u_{A,n} = (n+1) \frac{v}{2 l_A}$. The second overtone $(n=2)$ is $\

Vedclass · Accepted Answer

Both $(B)$ and $(C)$

Vedclass · Answer

(D) For an open organ pipe $A$,the frequency of the $n$-th overtone is given by $
u_{A,n} = (n+1) \frac{v}{2 l_A}$. The second overtone $(n=2)$ is $
u_{A,2} = \frac{3v}{2l_A}$.For a closed organ pipe $B$,the frequency of the $n$-th overtone is given by $
u_{B,n} = (2n+1) \frac{v}{4 l_B}$. The second overtone $(n=2)$ is $
u_{B,2} = \frac{5v}{4l_B}$.Given $
u_{A,2} = 
u_{B,2}$,we have $\frac{3v}{2l_A} = \frac{5v}{4l_B}$.Rearranging gives $\frac{l_B}{l_A} = \frac{5 	imes 2}{3 	imes 4} = \frac{10}{12} = \frac{5}{6}$. Thus,the ratio of lengths of $B$ to $A$ is $5:6$.Now,the first overtone of $A$ $(n=1)$ is $
u_{A,1} = \frac{2v}{2l_A} = \frac{v}{l_A}$.The first overtone of $B$ $(n=1)$ is $
u_{B,1} = \frac{3v}{4l_B}$.The ratio of these frequencies is $\frac{
u_{A,1}}{
u_{B,1}} = \frac{v}{l_A} 	imes \frac{4l_B}{3v} = \frac{4}{3} 	imes \frac{l_B}{l_A} = \frac{4}{3} 	imes \frac{5}{6} = \frac{20}{18} = \frac{10}{9}$.Both statements $(B)$ and $(C)$ are correct.

The second overtone of an open organ pipe $A$ and a closed pipe $B$ have the same frequency at a given temperature. It follows that the ratio of the

Similar Questions

$A$ tuning fork with frequency $800 \; Hz$ produces resonance in a resonance column tube with the upper end open and the lower end closed by a water surface. Successive resonances are observed at lengths $9.75 \; cm$,$31.25 \; cm$,and $52.75 \; cm$. The speed of sound in air is ...... $m/s$.

$A$ cylindrical tube open at both ends has a fundamental frequency '$f$' in air. The tube is dipped vertically in water so that half of its length is in water. The new fundamental frequency is

The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe of length $L$. The length of the open pipe will be

The fundamental frequency of an open pipe is $n$. If it is closed from one end, then the frequency of the $2^{nd}$ harmonic of the closed pipe is higher by $200 \,Hz$ than $n$. The value of $n$ is: (in $\,Hz$)

An open and a closed organ pipe are of same length. The ratio of the frequency of the $n^{th}$ mode of vibration of the two pipes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module