An open pipe of length L_1 has a fundamental | vedclass

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

Question

(C) For an open pipe of length $L_1$,the fundamental frequency is $n_1 = \frac{v}{2 L_1}$,which implies $L_1 = \frac{v}{2 n_1}$.For a closed pipe of l

Vedclass · Accepted Answer

$\frac{n_1 n_2}{n_1+2 n_2}$

Vedclass · Answer

(C) For an open pipe of length $L_1$,the fundamental frequency is $n_1 = \frac{v}{2 L_1}$,which implies $L_1 = \frac{v}{2 n_1}$.For a closed pipe of length $L_2$,the fundamental frequency is $n_2 = \frac{v}{4 L_2}$,which implies $L_2 = \frac{v}{4 n_2}$.The combined pipe is closed at one end and open at the other,with total length $L = L_1 + L_2$.The fundamental frequency of the combined pipe is $n = \frac{v}{4 L} = \frac{v}{4(L_1 + L_2)}$.Substituting the values of $L_1$ and $L_2$: $n = \frac{v}{4(\frac{v}{2 n_1} + \frac{v}{4 n_2})} = \frac{v}{4v(\frac{1}{2 n_1} + \frac{1}{4 n_2})} = \frac{1}{\frac{2}{4 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{1}{2 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{2 n_2 + n_1}{4 n_1 n_2}} = \frac{4 n_1 n_2}{n_1 + 2 n_2}$.Wait,re-evaluating the denominator: $n = \frac{1}{\frac{2}{4 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{1}{2 n_1} + \frac{1}{4 n_2}} = \frac{4 n_1 n_2}{2 n_2 + n_1}$.Given the options,the correct expression is $\frac{2 n_1 n_2}{n_1 + 2 n_2}$ if the length relations were different,but based on standard derivation,the result is $\frac{2 n_1 n_2}{n_1 + 2 n_2}$ is not matching. Let's re-check: $L = \frac{v}{2 n_1} + \frac{v}{4 n_2} = v \frac{2 n_2 + n_1}{4 n_1 n_2}$. Then $n = \frac{v}{4 L} = \frac{v}{4 v \frac{2 n_2 + n_1}{4 n_1 n_2}} = \frac{n_1 n_2}{n_1 + 2 n_2}$.

An open pipe of length $L_1$ has a fundamental frequency $n_1$,and a closed pipe of length $L_2$ has a fundamental frequency $n_2$. If they are joined to form a single closed pipe of length $L = L_1 + L_2$,what is the fundamental frequency of the combined pipe?

Similar Questions

$A$ tuning fork of frequency $n$ is held near the open end of a tube which is closed at the other end,and the length is adjusted until resonance occurs. The first resonance occurs at length $L_1$ and the immediate next resonance occurs at length $L_2$. The speed of sound in air is

In a resonance tube,the first resonance with a tuning fork occurs at $16 \ cm$ and the second at $49 \ cm$. If the velocity of sound is $330 \ m/s$,the frequency of the tuning fork is: (in $Hz$)

Two consecutive harmonics of an air column in a pipe closed at one end are of frequencies $150 \,Hz$ and $250 \,Hz$. The fundamental frequency of the air column is: (in $\,Hz$)

If the temperature increases,then what happens to the frequency of the sound produced by the organ pipe?

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 $\,cm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module