A closed organ pipe has a fundamental frequen | vedclass

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

Question

(A) The fundamental frequency of a closed organ pipe is given by $n = \frac{v}{4(L + 0.6r)}$,where $v$ is the speed of sound,$L$ is the length,and $r$

Vedclass · Accepted Answer

halved

Vedclass · Answer

(A) The fundamental frequency of a closed organ pipe is given by $n = \frac{v}{4(L + 0.6r)}$,where $v$ is the speed of sound,$L$ is the length,and $r$ is the radius.Since $L \gg r$,the end correction $0.6r$ is often neglected,simplifying the formula to $n \approx \frac{v}{4L}$.If the length $L$ is doubled $(L' = 2L)$ and the radius $r$ is halved $(r' = r/2)$,the new frequency $n'$ is given by:$n' = \frac{v}{4(L' + 0.6r')} = \frac{v}{4(2L + 0.6(r/2))} = \frac{v}{4(2L + 0.3r)}$.Since $L \gg r$,we can approximate $2L + 0.3r \approx 2L$.Therefore,$n' \approx \frac{v}{4(2L)} = \frac{1}{2} \left( \frac{v}{4L} \right) = \frac{n}{2}$.Thus,the frequency is halved.

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

Similar Questions

If the speed of sound in air is $330 \, m/s$,find the number of tones (harmonics) present in an open organ pipe of length $1 \, m$ whose frequency is $\leq 1000 \, Hz$.

An air column,closed at one end and open at the other,resonates with a tuning fork when the smallest length of the column is $50\, cm$. The next larger length of the column resonating with the same tuning fork is .... $cm$.

$A$ cylindrical tube $(L = 120\,cm)$ is resonant with a tuning fork of frequency $330\,Hz$. If it is being filled with water,what is the minimum length of the water column required to achieve resonance? (Speed of sound in air $V_{air} = 330\,m/s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module