Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 33

Using graph paper, sketch two identical sine waves of amplitude $4.0 \mathrm{cm}\( that differ in phase by (a) \)\pi / 3$ rad \(\left(60^{\circ}\right)\) and (b) \(\pi / 2\) rad \(\left(90^{\circ}\right) .\) Find the amplitude of the superposition of the two waves in each case.

Short Answer

Expert verified
Answer: The amplitudes of the superpositions for the two cases are approximately 6.93 cm and 5.66 cm, respectively.

Step by step solution

01

Sketch the first sine wave

Begin by drawing a sinusoidal curve that represents the first sine wave. This sine wave should have an amplitude of 4 cm.
02

Sketch the second sine wave (60° phase difference)

Now, draw another sine wave that has an amplitude of 4 cm and a phase difference of \(\pi / 3\) rad (60°) following the first sine wave.
03

Calculate the amplitude of the superposition

To find the amplitude of the superposition of the two sine waves, we can use the formula: \(A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}\), where \(A\) is the amplitude of the superposition, \(A_1\) and \(A_2\) are the amplitudes of the initial waves, and \(\Delta\phi\) is the phase difference between the waves. In our case, \(A_1 = A_2 = 4 \thinspace cm\), and \(\Delta\phi = \pi / 3\). So, \(A = \sqrt{4^2 + 4^2 + 2(4)(4) \cos(\pi / 3)} = \sqrt{48} \thinspace cm \approx 6.93 \thinspace cm\). #b) pi/2 (90 degrees) phase difference#
04

Sketch the sine wave (90° phase difference)

With the same amplitude of 4 cm, draw another sine wave that has a phase difference of \(\pi / 2\) rad (90°) from the first sine wave.
05

Calculate the amplitude of the superposition

Now we will use the same formula to find the amplitude of the superposition but with \(\Delta\phi = \pi / 2\): \(A = \sqrt{4^2 + 4^2 + 2(4)(4) \cos(\pi / 2)} = \sqrt{32} \thinspace cm \approx 5.66 \thinspace cm\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Deep-water waves are dispersive (their wave speed depends on the wavelength). The restoring force is provided by gravity. Using dimensional analysis, find out how the speed of deep-water waves depends on wavelength \(\lambda\), assuming that \(\lambda\) and \(g\) are the only relevant quantities. (Mass density does not enter into the expression because the restoring force, arising from the weight of the water, is itself proportional to the mass density.)
(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 \(\lambda=v T.\)
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 guitar string has a fundamental frequency of \(300.0 \mathrm{Hz}\) (a) What are the next three lowest standing wave frequencies? (b) If you press a finger lightly against the string at its midpoint so that both sides of the string can still vibrate, you create a node at the midpoint. What are the lowest four standing wave frequencies now? (c) If you press hard at the same point, only one side of the string can vibrate. What are the lowest four standing wave frequencies?
A standing wave has wave number $2.0 \times 10^{2} \mathrm{rad} / \mathrm{m} .$ What is the distance between two adjacent nodes?
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Physics of Motion

Read Explanation

Quantum Physics

Read Explanation

Kinematics Physics

Read Explanation

Rotational Dynamics

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.

Sign-up for free