In an open-end organ pipe of length L,if the | vedclass

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

Question

(A) For an open organ pipe of length $L$,the boundary conditions require antinodes at both open ends.The fundamental mode (first harmonic) corresponds

Vedclass · Accepted Answer

$\frac{V}{2L}$ and all harmonics are present.

Vedclass · Answer

(A) For an open organ pipe of length $L$,the boundary conditions require antinodes at both open ends.The fundamental mode (first harmonic) corresponds to the condition where the length of the pipe $L$ is equal to half the wavelength,i.e.,$L = \frac{\lambda}{2}$,so $\lambda = 2L$.Using the wave equation $V = f \lambda$,the fundamental frequency $f$ is given by $f = \frac{V}{\lambda} = \frac{V}{2L}$.In an open organ pipe,all harmonics (integer multiples of the fundamental frequency) are present,meaning both odd and even harmonics are produced.

In an open-end organ pipe of length $L$,if the velocity of sound is $V$,then the fundamental frequency will be (Neglect end correction).

Similar Questions

An organ pipe open at one end is vibrating in its first overtone and is in resonance with another pipe open at both ends vibrating in its third harmonic. The ratio of the length of the two pipes is:

$A$ cylindrical tube open at both ends has a fundamental frequency $f$ in air. When the tube is dipped vertically in water so that one-third part of the tube is in water,the fundamental frequency of the air column becomes (neglect end correction).

The waves set up in a closed pipe are

An organ pipe of length $L$ is open at one end and closed at the other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are

$A$ uniform narrow $1.95 \, m$ long pipe is open at both ends. It resonates at two successive harmonics of frequency $275 \, Hz$ and $330 \, Hz$. The speed of sound in the tube is ...... $m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module