/*! 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 5 Two waves are traveling in oppos... [FREE SOLUTION] | 91Ó°ÊÓ

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Two waves are traveling in opposite directions on the same string. The displacements caused by the individual waves are given by \(y_{1}=(24.0 \mathrm{mm}) \sin (9.00 \pi t-1.25 \pi x)\) and \(y_{2}=(35.0 \mathrm{mm}) \sin (2.88 \pi t+\) \(0.400 \pi x\) ). Note that the phase angles \((9.00 \pi t-1.25 \pi x)\) and \((2.88 \pi t+\) \(0.400 \pi x\) ) are in radians, \(t\) is in seconds, and \(x\) is in meters. At \(t=4.00 \mathrm{s}\) what is the net displacement (in \(\mathrm{mm}\) ) of the string at (a) \(x=2.16 \mathrm{m}\) and (b) \(x=2.56 \mathrm{m} ?\) Be sure to include the algebraic sign \((+\) or \(-)\) with your answers.

Short Answer

Expert verified
At \(x = 2.16\,\mathrm{m},\) the displacement is approximately \(\approx -37.4\,\text{mm}.\) At \(x = 2.56\,\mathrm{m},\) the displacement is \(\approx -1.69\,\text{mm}.\)

Step by step solution

01

Understand the Problem

We are given two wave equations, \( y_1 = (24.0 \ \text{mm}) \sin(9.00\pi t - 1.25\pi x) \) and \( y_2 = (35.0 \ \text{mm}) \sin(2.88\pi t + 0.400\pi x) \). The task is to find the net displacement \( y = y_1 + y_2 \) at two different positions \( x = 2.16 \ \text{m} \) and \( x = 2.56 \ \text{m} \) when \( t = 4.00 \ \text{s} \).
02

Calculate Individual Displacements at x = 2.16 m

Substitute \( t = 4.00 \ \text{s} \) and \( x = 2.16 \ \text{m} \) into the equations:1. For \( y_1 \): \[ y_1 = 24.0 \sin(9.00\pi \cdot 4.00 - 1.25\pi \cdot 2.16) \].2. For \( y_2 \):\[ y_2 = 35.0 \sin(2.88\pi \cdot 4.00 + 0.400\pi \cdot 2.16) \].Calculate these using a calculator, ensuring the angle calculations are in radians.
03

Calculate Net Displacement at x = 2.16 m

Compute the net displacement by adding the calculated values of \( y_1 \) and \( y_2 \) from Step 2:\[ y = y_1 + y_2 \].Ensure the addition considers the algebraic sign of each displacement.
04

Calculate Individual Displacements at x = 2.56 m

Substitute \( t = 4.00 \ \text{s} \) and \( x = 2.56 \ \text{m} \) into the wave equations:1. For \( y_1 \): \[ y_1 = 24.0 \sin(9.00\pi \cdot 4.00 - 1.25\pi \cdot 2.56) \].2. For \( y_2 \):\[ y_2 = 35.0 \sin(2.88\pi \cdot 4.00 + 0.400\pi \cdot 2.56) \].Use a calculator, ensuring angles are in radians.
05

Calculate Net Displacement at x = 2.56 m

Compute the net displacement by adding the calculated values of \( y_1 \) and \( y_2 \) from Step 4:\[ y = y_1 + y_2 \].Make sure to include the correct algebraic signs in your addition.

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Key Concepts

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

Superposition Principle
The superposition principle is a fundamental concept in physics, particularly in wave theory. It states that when two or more waves overlap, the resulting wave displacement at any point is the algebraic sum of the individual displacements caused by each wave. This principle ensures that waves can pass through each other without being permanently altered, merely summing their effects together at points where they meet. This makes it possible to predict the net effect of multiple waves on a medium, such as a string or an air column.

For our exercise, where two waves travel in opposite directions along a string, the superposition principle is used to find the net displacement of the string by simply adding the displacements from each wave, symbolized by \(y = y_1 + y_2\). When calculating this, it's crucial to keep track of the respective amplitudes and phase offsets of each wave, as these will dictate the behavior of the resulting displacement.

By applying the superposition principle, we can evaluate wave behaviors in various scenarios, such as constructive or destructive interference, which results from waves being in phase or out of phase, respectively. This makes the principle incredibly useful in understanding complex wave patterns seen in nature.
Wave Equations
Wave equations represent mathematical descriptions of wave forms as they propagate through a medium. In this context, the wave equations we are dealing with are for traveling waves along a string. These equation forms typically define the displacement of the medium in terms of sine or cosine functions, which are essential to modeling periodic wave behavior.

Each wave equation has several components:
  • An amplitude, which describes the maximum displacement of the wave.
  • A wave number, represented as an angular component (such as \(1.25\pi\) or \(0.400\pi\)), that conveys how many wave cycles fit into a unit length.
  • An angular frequency factor, such as \(9.00\pi\) or \(2.88\pi\), that relates to how many cycles occur within a time unit.
  • Phase angles that determine the initial phase of oscillation.
In our exercise example, the equations \(y_1 = (24.0\, ext{mm})\, \sin(9.00\pi t - 1.25\pi x)\) and \(y_2 = (35.0\, ext{mm})\, \sin(2.88\pi t + 0.400\pi x)\) help us determine how each wave produces displacement along the string over time. These wave equations are essential tools for both theoretical and practical wave analysis.
Phase Angles
Phase angles are crucial components of wave equations and play a significant role in how waves interact. In sinusoidal functions, such as the ones present in our problem, phase angles signify the specific point in the cycle of a wave at any given instant. This is key in understanding and predicting wave interference effects.

The phase angle of a wave is influenced by:
  • The term involving time \( t \) (e.g., \(9.00\pi t\) or \(2.88\pi t\)), which determines the effect of frequency on the wave's phase.
  • The spatial term \( x \) (e.g., \(-1.25\pi x\) or \(+0.400\pi x\)), which accounts for the wave's position in space.
In practice, when two waves combine, the relative phase angles determine whether they will interfere constructively or destructively:
  • Constructive interference occurs when the waves are in phase (their crests and troughs align), leading to larger displacement.
  • Destructive interference happens when the waves are out of phase (crests align with troughs of another), potentially even canceling each other out.
In the given exercise, examining the phase angles helps us calculate the precise point in space and time where these waves will interact, allowing us to find the resulting displacement of the string at specific positions.

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Most popular questions from this chapter

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube open at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about \(2.9 \mathrm{cm} .\) The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the fundamental frequency of the canal? (Interestingly, the fundamental frequency is in the frequency range where human hearing is most sensitive.)

Multiple-Concept Example 3 reviews the concepts that are important in this problem. The entrance to a large lecture room consists of two side-by-side doors, one hinged on the left and the other hinged on the right. Each door is \(0.700 \mathrm{m}\) wide. Sound of frequency \(607 \mathrm{Hz}\) is coming through the entrance from within the room. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the diffraction angle \(\theta\) of the sound after it passes through the doorway when (a) one door is open and (b) both doors are open?

A vertical tube is closed at one end and open to air at the other end. The air pressure is \(1.01 \times 10^{5} \mathrm{Pa}\). The tube has a length of \(0.75 \mathrm{m} .\) Mercury (mass density \(\left.=13600 \mathrm{kg} / \mathrm{m}^{3}\right)\) is poured into it to shorten the effective length for standing waves. What is the absolute pressure at the bottom of the mercury column, when the fundamental frequency of the shortened, air-filled tube is equal to the third harmonic of the original tube?

A tube, open at only one end, is cut into two shorter (nonequal) lengths. The piece that is open at both ends has a fundamental frequency of \(425 \mathrm{Hz},\) while the piece open only at one end has a fundamental frequency of \(675 \mathrm{Hz}\). What is the fundamental frequency of the original tube?

To review the concepts that play roles in this problem, consult Multiple- Concept Example \(4 .\) Sometimes, when the wind blows across a long wire, a low-frequency "moaning" sound is produced. This sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density \(=\) \(0.0140 \mathrm{kg} / \mathrm{m}\) ) sustains a tension of \(323 \mathrm{N}\) because the wire is stretched between two poles that are \(7.60 \mathrm{m}\) apart. The lowest frequency that an average, healthy human ear can detect is \(20.0 \mathrm{Hz}\). What is the lowest harmonic number \(n\) that could be responsible for the "moaning" sound?

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