/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 Two waves are described by $$ ... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 89

Two waves are described by $$ y_{1}=0.30 \sin [\pi(5 x-200) t] $$ and $$ y_{2}=0.30 \sin [\pi(5 x-200 t)+\pi / 3] $$ where \(y_{1}, y_{2}\), and \(x\) are in meters and \(t\) is in seconds. When these two waves are combined, a traveling wave is produced. What are the (a) amplitude, (b) wave speed, and (c) wavelength of that traveling wave?

Short Answer

Expert verified
Amplitude: 0.30 m, Wave speed: 40 m/s, Wavelength: 0.4 m.

Step by step solution

01

Identify the Components of Each Wave

The given equations for the waves are:\( y_{1} = 0.30 \sin[\pi(5x - 200t)] \) and \( y_{2} = 0.30 \sin[\pi(5x - 200t) + \pi/3] \).Both waves have the same amplitude of 0.30 m, same wave number \( k = \pi \times 5 \), and angular frequency \( \omega = \pi \times 200 \).
02

Use the Principle of Superposition

The superposition principle states that the resultant wave \( y \) is the sum of the two waves:\[ y = y_{1} + y_{2} = 0.30 \sin[\pi(5x - 200t)] + 0.30 \sin[\pi(5x - 200t) + \frac{\pi}{3}] \]These two sine functions can be combined into a single wave using trigonometric identities.
03

Apply Trigonometric Identity

Use the identity for the sum of sine functions, \( \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \).Let \( A = \pi(5x - 200t) \) and \( B = \pi(5x - 200t) + \pi/3 \). The expression becomes:\[ y = 2 \times 0.30 \sin\left(\pi(5x - 200t) + \frac{\pi}{6}\right) \cos\left(\frac{\pi}{3}\right) \]Since \( \cos(\pi/3) = 0.5 \), the resultant amplitude is:\[ 0.30 \times 2 \times 0.5 = 0.30 \]
04

Calculate the Wave Speed

The wave speed \( v \) can be found using relation \( v = \frac{\omega}{k} \), where \( \omega = \pi \times 200 \) and \( k = \pi \times 5 \).\[ v = \frac{\pi \times 200}{\pi \times 5} = \frac{200}{5} = 40 \text{ m/s} \]
05

Determine the Wavelength

The wave number \( k \) is related to the wavelength \( \lambda \) by \( k = \frac{2\pi}{\lambda} \). We have \( k = \pi \times 5 \).\[ \pi \times 5 = \frac{2\pi}{\lambda} \Rightarrow \lambda = \frac{2\pi}{5\pi} = \frac{2}{5} \text{ meters} \]
06

Conclusion

The amplitude of the resultant wave is 0.30 meters, the wave speed is 40 meters per second, and the wavelength is 0.4 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude Calculation
When dealing with wave superposition, calculating the amplitude of the resulting wave is a crucial step. In this exercise, two waves with identical amplitudes of 0.30 meters are combined, utilizing the principle of superposition. The superposition principle allows us to sum up the amplitudes of individual waves mathematically.
By applying the trigonometric identity for the sum of sine functions, we can combine the two waves into a single wave. The identity used here is:- \( \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \)
After simplification, the cosine function impacts the total amplitude, effectively multiplying it by the cosine of the phase difference, \( \cos(\pi/3) = 0.5 \). Consequently, the resultant wave amplitude is:- \( 0.30 \times 2 \times 0.5 = 0.30 \) meters
Surprisingly, the initial and final amplitudes remain the same due to the specific phase difference, illustrating how phase differences can impact wave results in superposition.
Wave Speed
Wave speed, symbolized by \( v \), indicates how quickly a wave propagates through a medium. We can find this speed using the equation:- \( v = \frac{\omega}{k} \)
Here, \( \omega \) is the angular frequency, and \( k \) is the wave number. Both are provided for the waves in the exercise:- Angular frequency \( \omega = \pi \times 200 \)- Wave number \( k = \pi \times 5 \)
By substituting these values into the equation, we obtain:- \( v = \frac{\pi \times 200}{\pi \times 5} = \frac{200}{5} = 40 \text{ m/s} \)
This calculation shows that the wave travels at a speed of 40 meters per second, reflecting how distance and time relate within wave movements.
Wavelength Determination
Determining the wavelength involves a clear understanding of the relationship between the wave number \( k \) and the wavelength \( \lambda \). The relationship is governed by the equation:- \( k = \frac{2\pi}{\lambda} \)
Given in the exercise, the wave number is \( \pi \times 5 \). Using this, we can determine the wavelength as follows:- \[ \pi \times 5 = \frac{2\pi}{\lambda} \]
This equation can be simplified to solve for \( \lambda \):- \[ \lambda = \frac{2\pi}{5\pi} = \frac{2}{5} \text{ meters} \]
Thus, the wavelength of the wave is 0.4 meters. This length represents the distance between consecutive peaks (or troughs) of the wave, illustrating how spatial characteristics of a wave are quantified.

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Most popular questions from this chapter

A sinusoidal wave is traveling on a string with speed \(40 \mathrm{~cm} / \mathrm{s}\). The displacement of the particles of the string at \(x=10 \mathrm{~cm}\) varies with time according to \(y=(5.0 \mathrm{~cm}) \sin \left[1.0-\left(4.0 \mathrm{~s}^{-1}\right) t\right]\). The linear density of the string is \(4.0 \mathrm{~g} / \mathrm{cm}\). What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (c) \(y_{m},(\mathrm{~d}) k,(\mathrm{e}) \omega\), and (f) the correct choice of sign in front of \(\omega ?(\mathrm{~g})\) What is the tension in the string?

The type of rubber band used inside some baseballs and golf balls obeys Hooke's law over a wide range of elongation of the band. A segment of this material has an unstretched length \(\ell\) and a mass \(m\). When a force \(F\) is applied, the band stretches an additional length \(\Delta \ell\). (a) What is the speed (in terms of \(m, \Delta \ell\), and the spring constant \(k\) ) of transverse waves on this stretched rubber band? (b) Using your answer to (a), show that the time required for a transverse pulse to travel the length of the rubber band is proportional to \(1 / \sqrt{\Delta \ell}\) if \(\Delta \ell \ll \ell\) and is constant if \(\Delta \ell \geqslant \ell\).

The equation of a transverse wave traveling along a very long string is \(y=6.0 \sin (0.020 \pi x+4.0 \pi t)\), where \(x\) and \(y\) are expressed in centimeters and \(t\) is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. \((\mathrm{g})\) What is the transverse displacement at \(x=3.5 \mathrm{~cm}\) when \(t=0.26 \mathrm{~s}\) ?

(a) What is the fastest transverse wave that can be sent along a steel wire? For safety reasons, the maximum tensile stress to which steel wires should be subjected is \(7.00 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\). The density of steel is \(7800 \mathrm{~kg} / \mathrm{m}^{3}\). (b) Does your answer depend on the diameter of the wire?

If If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind. The air pressure variations in the vortexes tend to cause the line to oscillate (gallop), especially if the frequency of the variations matches a resonant frequency of the line. In long lines, the resonant frequencies are so close that almost any wind speed can set up a resonant mode vigorous enough to pull down support towers or cause the line to short out with an adjacent line. If a transmission line has a length of \(347 \mathrm{~m}\), a linear density of \(3.35 \mathrm{~kg} / \mathrm{m}\), and a tension of \(65.2 \mathrm{MN}\), what are (a) the frequency of the fundamental mode and (b) the frequency difference between successive modes?
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