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Chapter 10: Problem 1487

Which of the following changes at an antinode in a stationary wave? (A) Density only (B) Pressure only (C) Both pressure and density (D) Neither pressure nor density

Short Answer

Expert verified
At an antinode in a stationary wave, particles oscillate with maximum amplitude, causing them to become compressed or rarefied at different moments of their oscillation. This results in changes in both density and pressure. Therefore, the correct answer is (C) Both pressure and density.

Step by step solution

01

Understand stationary waves and antinodes

A stationary wave, also known as a standing wave, is formed when two waves of the same frequency, amplitude, and wavelength traveling in opposite directions interfere with each other. An antinode is the position in a stationary wave where the amplitude of oscillation is maximum. That means the particles at an antinode have the maximum displacement.
02

Analyze the properties of antinodes

The density of a medium is defined as its mass per unit volume, whereas pressure is defined as a force exerted per unit area. At an antinode, particles oscillate with maximum amplitude, which means that their displacement from their equilibrium position is maximum. Because of this displacement, the particles can become compressed or rarefied at different moments of their oscillation.
03

Determine the change in density and pressure at an antinode

When the particles of the medium are compressed, they are closer together, and therefore, the density is higher, and so is the pressure. In contrast, when the particles are rarefied, they are farther apart, leading to lower density and pressure. Since the particles oscillate between being compressed and rarefied, both density and pressure change at an antinode in a stationary wave.
04

Choose the correct option

Based on the analysis, we can conclude that both density and pressure change at an antinode in a stationary wave. Therefore, the correct answer is: (C) Both pressure and density

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Most popular questions from this chapter

For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. (a) Statement \(-1\) is true, statement \(-2\) is true; statement \(-2\) is the correct explanation of statement \(-1\). (b) Statement \(-1\) is true, statement \(-2\) is true but statement \(-2\) is not the correct explanation of statement \(-1\) (c) Statement \(-1\) is true, statement \(-2\) is false (d) Statement \(-1\) is false, statement \(-2\) is true (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) d Statement \(-1:\) For a particle executing S.H.M. with an amplitude of $0.01 \mathrm{~m}\( and frequency \)30 \mathrm{hz}\(, the maximum acceleration is \)36 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}$. Statement \(-2:\) The maximum acceleration for the above particle is $\pm \omega 2 \mathrm{~A}\(, where \)\mathrm{A}$ is amplitude. (A) a (B) \(\mathrm{b}\) (C) \(c\) (D) \(\mathrm{d}\)

The speed of a particle executing motion changes with time according to the equation $\mathrm{y}=\mathrm{a} \sin \omega \mathrm{t}+\mathrm{b} \cos \omega \mathrm{t}\(, then \)\ldots \ldots \ldots$ (A) Motion is periodic but not a S.H.M. (B) It is a S.H.M. with amplitude equal to \(\mathrm{a}+\mathrm{b}\) (C) It is a S.H.M. with amplitude equal to \(\mathrm{a}^{2}+\mathrm{b}^{2}\)

A metal wire having linear mass density \(10 \mathrm{~g} / \mathrm{m}\) is passed over two supports separated by a distance of \(1 \mathrm{~m}\). The wire is kept in tension by suspending a \(10 \mathrm{~kg}\) mass. The mid point of the wire passes through a magnetic field provided by magnets and an a. c. supply having frequency \(\mathrm{n}\) is passed through the wire. If the wire starts vibrating with its resonant frequency, what is the frequency of a. c. supply? (A) \(50 \mathrm{~Hz}\) (B) \(100 \mathrm{~Hz}\) (C) \(200 \mathrm{~Hz}\) (D) \(25 \mathrm{~Hz}\)

A wire having length \(\mathrm{L}\) is kept under tension between \(\mathrm{x}=0\) and \(\mathrm{x}=\mathrm{L}\). In one experiment, the equation of the wave and energy is given by $\mathrm{y}_{1}=\mathrm{A} \sin (\pi \mathrm{x} / \mathrm{L}) \sin \omega \mathrm{t}\( and \)\mathrm{E}_{1}$ respectively. In another experiment, it is \(\mathrm{y}_{2}=\mathrm{A} \sin\) \(\\{(2 \pi \mathrm{x}) / \mathrm{L}\\} \sin 2 \omega \mathrm{t}\) and \(\mathrm{E}_{2}\). Then........ (A) \(E_{2}=E_{1}\) (B) \(E_{2}=2 \mathrm{E}_{1}\) (C) \(\mathrm{E}_{2}=4 \mathrm{E}_{1}\) (D) \(E_{2}=16 \mathrm{E}_{1}\)

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.
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