/*! 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 35 A traveling sine wave is the res... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 35

A traveling sine wave is the result of the superposition of two other sine waves with equal amplitudes, wavelengths, and frequencies. The two component waves each have amplitude \(5.00 \mathrm{cm} .\) If the superposition wave has amplitude \(6.69 \mathrm{cm},\) what is the phase difference \(\phi\) between the component waves? [Hint: Let \(y_{1}=A \sin (\omega t+k x)\) and $y_{2}=A \sin (\omega t+k x-\phi) .$ Make use of the trigonometric identity (Appendix A.7) for \(\sin \alpha+\sin \beta\) when finding \(y=y_{1}+y_{2}\) and identify the new amplitude in terms of the original amplitude. \(]\)

Short Answer

Expert verified
Answer: The phase difference between the two component waves is approximately 96.4 degrees.

Step by step solution

01

Write the given equations

The given equations for the component waves are: \(y_1 = A \sin (\omega t + kx)\) and \(y_2 = A \sin (\omega t + kx - \phi)\) where A is the amplitude, k is the wave number, \(\omega\) is the angular frequency, and \(\phi\) is the phase difference between the two waves.
02

Add the component waves to find the resultant wave

To find the resultant wave, we need to add the component waves: \(y = y_1 + y_2\) \(y = A\sin(\omega t + kx) + A\sin(\omega t + kx - \phi)\)
03

Use the trigonometric identity for adding sine functions

We use the trigonometric identity for adding sine functions: \(\sin{\alpha}+\sin{\beta}=2\sin{\frac{\alpha + \beta}{2}}\cos{\frac{\alpha - \beta}{2}}\) Applying this identity to the equation for the resultant wave: \(y = 2A\sin{\frac{(\omega t + kx) + (\omega t + kx - \phi)}{2}}\cos{\frac{(\omega t + kx) - (\omega t + kx - \phi)}{2}}\) Simplify the equation: \(y = 2A\sin({\omega t + kx - \frac{\phi}{2}})\cos{-\frac{\phi}{2}}\)
04

Write amplitude relationship and solve for 蠁

We are given that the amplitude of the resultant wave is 6.69 cm. Therefore, the amplitude of the superposition wave can be written as: \(6.69 \mathrm{cm} = 2(5.00 \mathrm{cm})\cos{-\frac{\phi}{2}}\) Solve for \(\cos{-\frac{\phi}{2}}\): \(\cos{-\frac{\phi}{2}} = \frac{6.69 \mathrm{cm}}{2(5.00 \mathrm{cm})} = 0.669\) Now solve for the phase difference 蠁: \(\phi = -2 \cos^{-1}(0.669)\) \(\phi \approx 2(48.2^\circ)\) \(\phi \approx 96.4^\circ\) The phase difference between the two component waves is approximately 96.4 degrees.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the amplitudes of the graphs you made in Problem 74 satisfy the equation \(A^{\prime}=2 A \cos (\omega t),\) where \(A^{\prime}\) is the amplitude of the wave you plotted and \(A\) is \(5.0 \mathrm{cm},\) the amplitude of the waves that were added together.
Suppose that a string of length \(L\) and mass \(m\) is under tension \(F\). (a) Show that \(\sqrt{F L} m\) has units of speed. (b) Show that there is no other combination of \(L, m,\) and \(F\) with units of speed. [Hint: Of the dimensions of the three quantities $L, m, \text { and } F, \text { only } F \text { includes time. }]$ Thus, the speed of transverse waves on the string can only be some dimensionless constant times \(\sqrt{F L / m}.\)
(a) Sketch graphs of \(y\) versus \(x\) for the function $$ y(x, t)=(0.80 \mathrm{mm}) \sin (k x-\omega t) $$ for the times \(t=0,0.96 \mathrm{s},\) and \(1.92 \mathrm{s} .\) Make all three graphs of the same axes, using a solid line for the first, a dashed line for the second, and a dotted line for the third. Use the values \(k=\pi /(5.0 \mathrm{cm})\) and $\omega=(\pi / 6.0) \mathrm{rad} / \mathrm{s}$ (b) Repeat part (a) for the function $$ y(x, t)=(0.50 \mathrm{mm}) \sin (k x+\omega t) $$ (c) Which function represents a wave traveling in the \(-x\) direction and which represents a wave traveling in the \(+x\) -direction?
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) Plot a graph for $$ y(x, t)=(4.0 \mathrm{cm}) \sin [(378 \mathrm{rad} / \mathrm{s}) t-(314 \mathrm{rad} / \mathrm{cm}) x] $$ versus \(x\) at \(t=0\) and at \(t=\frac{1}{480} \mathrm{s}\). From the plots determine the amplitude, wavelength, and speed of the wave. (b) For the same function, plot a graph of \(y(x, t)\) versus \(t\) at \(x=0\) and find the period of the vibration. Show that \(\lambda=v T.\)
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Astrophysics

Read Explanation

Measurements

Read Explanation

Electricity

Read Explanation

Physics of Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.