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Chapter 11: Problem 47

A standing wave has wave number $2.0 \times 10^{2} \mathrm{rad} / \mathrm{m} .$ What is the distance between two adjacent nodes?

Short Answer

Expert verified
Answer: The distance between two adjacent nodes in the standing wave is 0.0157 meters.

Step by step solution

01

Determine the wavelength of the wave

We are given the wave number k = \(2.0 \times 10^{2} \mathrm{rad} / \mathrm{m}\). To find the wavelength, we can use the expression \(k = \frac{2\pi}{λ}\). Rearrange the equation to solve for λ: \(λ = \frac{2\pi}{k}\) Now, plug in the given value of k: \(λ = \frac{2\pi}{2.0 \times 10^{2} \mathrm{rad} / \mathrm{m}}\)
02

Evaluate the expression for the wavelength

Calculate the value for the wavelength: \(λ = \frac{2\pi}{2.0 \times 10^{2} \mathrm{rad} / \mathrm{m}} \approx \frac{6.28}{200} \approx 0.0314 \mathrm{m}\)
03

Find the distance between adjacent nodes

The distance between adjacent nodes in a standing wave is half the wavelength. Therefore, the distance between two adjacent nodes can be found by dividing the wavelength by 2: Distance between adjacent nodes = \(\frac{λ}{2}\) Now plug in the value we found for the wavelength: Distance between adjacent nodes = \(\frac{0.0314 \mathrm{m}}{2}\)
04

Calculate the distance between adjacent nodes

Evaluate the expression for the distance between adjacent nodes: Distance between adjacent nodes = \(\frac{0.0314 \mathrm{m}}{2} = 0.0157 \mathrm{m}\) The distance between two adjacent nodes in the standing wave is 0.0157 meters.

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Most popular questions from this chapter

A transverse wave on a string is described by $y(x, t)=(1.2 \mathrm{mm}) \sin [(2.0 \pi \mathrm{rad} / \mathrm{s}) t-(0.50 \pi \mathrm{rad} / \mathrm{m}) x]$ Plot the displacement \(y\) and the velocity \(v_{y}\) versus \(t\) for one complete cycle of the point \(x=0\) on the string.
A 1.6 -m-long string fixed at both ends vibrates at resonant frequencies of \(780 \mathrm{Hz}\) and \(1040 \mathrm{Hz}\), with no other resonant frequency between these values. (a) What is the fundamental frequency of this string? (b) When the tension in the string is \(1200 \mathrm{N},\) what is the total mass of the string?
Two waves with identical frequency but different amplitudes $A_{1}=6.0 \mathrm{cm}\( and \)A_{2}=3.0 \mathrm{cm},$ occupy the same region of space (i.e., are superimposed). (a) At what phase difference will the resulting wave have the highest intensity? What is the amplitude of the resulting wave in that case? (b) At what phase difference will the resulting wave have the lowest intensity and what will its amplitude be? (c) What is the ratio of the two intensities?
A transverse wave on a string is described by the equation $y(x, t)=(2.20 \mathrm{cm}) \sin [(130 \mathrm{rad} / \mathrm{s}) t+(15 \mathrm{rad} / \mathrm{m}) x].$ (a) What is the maximum transverse speed of a point on the string? (b) What is the maximum transverse acceleration of a point on the string? (c) How fast does the wave move along the string? (d) Why is your answer to (c) different from the answer to (a)?
A uniform string of length \(10.0 \mathrm{m}\) and weight \(0.25 \mathrm{N}\) is attached to the ceiling. A weight of \(1.00 \mathrm{kN}\) hangs from its lower end. The lower end of the string is suddenly displaced horizontally. How long does it take the resulting wave pulse to travel to the upper end? [Hint: Is the weight of the string negligible in comparison with that of the hanging mass?]
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