**Question:**A standing wave vibrating at the 2nd harmonic is produced in a tube of length 42 cm which is closed at one of its ends. What is the wave length of the standing wave in meters?**Explanation:**For a tube closed at one end, the second harmonic (also known as the first overtone) forms with a pattern similar to 3/4 of the wavelength fitting into the length of the tube. Therefore, the length of the tube $ L $ is equal to $ \frac{3}{4} \lambda $, where $ \lambda $ is the wavelength.Given:- Length of the tube $ L = 42 $ cm = $ 0.42 $ metersTo find the wavelength $ \lambda $, use the relationship:\[ \frac{3}{4} \lambda = L \]Rearrange to solve for $ \lambda $:\[ \lambda = \frac{4}{3} L \]Substituting the given length:\[ \lambda = \frac{4}{3} \times 0.42 \]\[ \lambda = 0.56 \text{ meters} \]Therefore, the wavelength of the standing wave is 0.56 meters.

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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For a pipe closed at one end and open at the other a node forms at the closed end and an antinode forms at the open end. The node is the position of least displacement and the antinode represents the position for maximum displacement . If we consider $λ$ to be the wavelength of the standing wave and $L$ be the length of the pipe then ,

$L = \frac{λ}{4}$

This due to the fact we are considering the distance between a node and an antinode. If this would have been a distance between two antinodes or nodes, then, $λ = 2 L$ .

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