Question

Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $\mathrm{A}$ and $\mathrm{B}$ is $\ldots \ldots \ldots$ (A) $1: 2$ (B) $1: 4$ (C) $2: 1$ (D) $4: 1$

Name: Problem 1465, Chapter 10: JEE Physics
Uploaded: 2021-11-12T16:50:08-08:00
Duration: 2 min 11 s
Channel: Akshaya Rs
Description: Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube A and B is ………(A) 1: 2 (B) 1: 4 (C) 2: 1 (D) 4: 1

   Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $\mathrm{A}$ and $\mathrm{B}$ is $\ldots \ldots \ldots$
(A) $1: 2$
(B) $1: 4$
(C) $2: 1$
(D) $4: 1$

JEE Physics

StemEZ 1st Edition

Chapter 10, Problem 1465

Instant Answer

Solved by Expert Akshaya Rs

11/12/2021

Step 1

Step 1: The fundamental frequency of a tube is given by the formula $f = \frac{v}{2L}$ for a tube with both ends open and $f = \frac{v}{4L}$ for a tube with one end closed, where $v$ is the speed of sound and $L$ is the length of the tube. Show more…

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Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $\mathrm{A}$ and $\mathrm{B}$ is $\ldots \ldots \ldots$ (A) $1: 2$ (B) $1: 4$ (C) $2: 1$ (D) $4: 1$

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Key Concepts

Standing Waves in Pipes

Standing waves are formed in tubes when incident waves reflect off boundaries and interfere with incoming waves. The conditions at the ends of the tube (whether open or closed) dictate the formation of nodes (points of minimal displacement) and antinodes (points of maximal displacement), which in turn determine the possible wavelengths that can exist in the tube.

Boundary Conditions in Open and Closed Tubes

In an open tube, both ends support displacement antinodes, resulting in a fundamental mode where the wavelength is twice the length of the tube (? = 2L) and the frequency is given by v/(2L). In contrast, a tube closed at one end forces a displacement node at the closed end and an antinode at the open end, producing a fundamental wavelength of four times the tube's length (? = 4L) and a corresponding frequency of v/(4L).

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