An open organ pipe having fundamental frequen | vedclass

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

Question

(B) The fundamental frequency of an open organ pipe of length $L$ is given by $n_{	ext{open}} = \frac{v}{2L} = n$ ... $(i)$ Since the pipe is in unis

Vedclass · Accepted Answer

$2: 1$

Vedclass · Answer

(B) The fundamental frequency of an open organ pipe of length $L$ is given by $n_{	ext{open}} = \frac{v}{2L} = n$ ... $(i)$ Since the pipe is in unison with the string,the frequency of the string is $n_s = n = \frac{v}{2L}$. When the tube is dipped in water such that $75 \%$ of its length is inside,the remaining length of the air column is $l_1 = 25 \% 	imes L = \frac{L}{4}$. The pipe now acts as a closed organ pipe (closed at one end). The fundamental frequency of this closed pipe is $n_{	ext{closed}} = \frac{v}{4l_1} = \frac{v}{4(L/4)} = \frac{v}{L}$. We need the ratio of the fundamental frequency of the air column of the dipped tube $(n_{	ext{closed}})$ to that of the string $(n_s)$: Ratio $= \frac{n_{	ext{closed}}}{n_s} = \frac{v/L}{v/2L} = \frac{2}{1}$.

Similar Questions

What should be the length of a closed pipe to produce resonance with a sound wave of wavelength $62 \,cm$,in fundamental mode (in $\,cm$)? [Neglect end correction]

The number of possible natural oscillations of an air column in a pipe closed at one end of length $85 \, cm$ whose frequencies lie below $1250 \, Hz$ are (Velocity of sound $= 340 \, m s^{-1}$)

In a closed organ pipe, the number of nodes formed in the fifth and ninth harmonics are respectively:

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

Two closed organ pipes having lengths $20\,cm$ and $20.5\,cm$ produce $5\,beats/sec$. Determine the frequency of both organ pipes.

Vedclass Test Series

Exam Paper Generator

Online Exam Module