/*! 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 70 Two in-phase loudspeakers are lo... [FREE SOLUTION] | 91Ó°ÊÓ

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Two in-phase loudspeakers are located at \((x, y)\) coordinates \((-3.0 \mathrm{m},+2.0 \mathrm{m})\) and \((-3.0 \mathrm{m},-2.0 \mathrm{m}) .\) They emit identical sound waves with a \(2.0 \mathrm{m}\) wavelength and amplitude \(a\). Determine the amplitude of the sound at the five positions on the \(y\) axis \((x=0)\) with \(y=0.0 \mathrm{m}, 0.5 \mathrm{m}, 1.0 \mathrm{m}, 1.5 \mathrm{m},\) and \(2.0 \mathrm{m}\).

Short Answer

Expert verified
The amplitude of the sound at the five positions on the y axis are obtained using the superposition principle and the phase differences calculated. The amplitude values would depend on the phase differences and individual amplitude of the sound wave.

Step by step solution

01

Calculate the Distance from Each Speaker to the Point of Interest

Use Pythagoras' theorem to calculate the distance \(d\) from each speaker to each point of interest. The formula to use is \(d = \sqrt{x^2 + y^2}\) where \(x = 3.0 m\) and \(y = 0.0 m, 0.5 m, 1.0 m, 1.5 m, 2.0 m\) respectively.
02

Determine the Phase Difference for Each Speaker

For each point of interest, calculate the phase difference \(\Delta \phi\) for each speaker. This is given by \(\Delta \phi = \frac{2 \pi d}{\lambda}\) where \(\lambda = 2.0 m\) is the wavelength and \(d\) is the distance from each speaker to the point of interest.
03

Apply the Principle of Superposition

To determine the resulting amplitude at each point of interest, apply the principle of superposition. The total displacement (y_total) at a point is given by \(y_{total}=a\sqrt{2(1+cos(\Delta \phi))}\) where \(a\) is the amplitude of each individual wave and \(\Delta \phi\) is the phase difference. Compute this for each point on the y-axis.

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Key Concepts

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

Principle of Superposition
The concept known as the Principle of Superposition is fundamental in understanding how waves interact. It states that when two or more waves meet, the total displacement at any point is the sum of the displacements of the individual waves at that point. This principle applies to various types of waves, including sound waves, light waves, and water waves.

For example, if two sound waves with identical frequencies and amplitudes intersect, the air particles at the point of intersection will simultaneously experience the push and pull from both waves. As a result, the displacements caused by the individual waves are added together to produce a new resultant wave that has a different amplitude than the original waves.

Understandably, if the waves are in-sync or in-phase, they will constructively interfere, reinforcing each other to create a wave of greater amplitude. Conversely, if they are out of phase, they can destructively interfere, leading to reduced amplitude or even cancellation.
Wavelength Calculation
Wavelength is a key concept in the study of wave phenomena and is denoted by the Greek letter lambda \( \lambda \). It represents the distance between consecutive points of medium displacement that are in phase, such as the distance between two adjacent crests or troughs in a wave.

In the context of the textbook problem, the wavelength given for the sound waves is \(2.0 \mathrm{m}\). This value remains constant for a specific frequency and medium and is crucial for calculating various properties of the wave, including the phase difference between two points in space.

To calculate wavelength in other scenarios, one would typically need to know the frequency of the wave and the speed at which it travels, using the formula: \[ \lambda = \frac{v}{f} \] where \(v\) is the speed of the wave and \(f\) is its frequency. In our case, this is not necessary, as the wavelength is provided.
Phase Difference
The Phase Difference is the relative displacement between the positions of two waves that are oscillating at the same frequency, usually measured in radians. This phase difference, denoted as \( \Delta \phi \), determines the nature of interference (constructive or destructive) between waves.

In the case of sound waves from the two loudspeakers in the exercise, knowing the phase difference helps to predict how the waves will combine at a given point. It's calculated by the formula: \[ \Delta \phi = \frac{2\pi d}{\lambda} \] where \(d\) is the distance the wave has traveled and \(\lambda\) is the wavelength.

If the resulting \( \Delta \phi \) is a multiple of \(2\pi\), the waves reinforce each other, causing constructive interference. But if the phase difference is an odd multiple of \(\pi\), the waves will tend to cancel each other out, leading to destructive interference. Calculating these differences is essential in predicting the behavior of waves at various points in space.
Pythagoras' Theorem in Physics
In the realm of physics, Pythagoras' Theorem is often employed to determine the distance between two points in space, which is a prerequisite for many wave-related calculations. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

This concept is critical for sound wave problems like the one in our exercise. To elucidate, consider each speaker and the point of interest forming a right-angled triangle with the horizontal and vertical distances representing the two shorter sides. The theorem is mathematically expressed as: \[ d = \sqrt{x^2 + y^2} \] where \(d\) is the distance from the speaker to the point of interest (the hypotenuse), \(x\) is the horizontal distance, and \(y\) is the vertical distance.

Employing this theorem allows us to calculate the distances required to then determine the phase difference for waves emanating from two different sources, setting the stage for analyzing their interference pattern.

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Most popular questions from this chapter

A vertical tube, open at both ends, is lowered into a tank of water until it is partially filled. The top portion of the tube, above the water, is filled with a gas that, because it is denser than air, remains in the tube. A 50.0 -cm-long, 1.00 g horizontal wire is stretched just above the top of the tube with \(440 \mathrm{N}\) of tension. Bowing the wire at its center causes the wire to vibrate at its fundamental frequency. The water level in the tube is adjusted until the sound from the vibrating wire sets up a standing sound wave in the gas. The water is then lowered another \(30.5 \mathrm{cm}\) until the next standing sound wave is detected. Use this information to determine the speed of sound in the gas.

Two sinusoidal waves with equal wavelengths travel along a string in opposite directions at \(3.0 \mathrm{m} / \mathrm{s}\). The time between two successive instants when the antinodes are at maximum height is \(0.25 \mathrm{s} .\) What is the wavelength?

A heavy piece of hanging sculpture is suspended by a \(90-\mathrm{cm}-\) long, \(5.0 \mathrm{g}\) steel wire. When the wind blows hard, the wire hums at its fundamental frequency of \(80 \mathrm{Hz}\). What is the mass of the sculpture?

a. The frequency of a standing wave on a string is \(f\) when the string's tension is \(T\). If the tension is changed by the small amount \(\Delta T\), without changing the length, show that the frequency changes by an amount \(\Delta f\) such that 1 $$ \frac{\Delta f}{f}=\frac{1}{2} \frac{\Delta T}{T} $$ b. Two identical strings vibrate at \(500 \mathrm{Hz}\) when stretched with the same tension. What percentage increase in the tension of one of the strings will cause five beats per second when both strings vibrate simultancously?

A water wave is called a deep-water wave if the water's depth is more than one-quarter of the wavelength. Unlike the waves we've considered in this chapter, the speed of a deep-water wave depends on its wavelength: $$ v=\sqrt{\frac{g \lambda}{2 \pi}} $$ Longer wavelengths travel faster. Let's apply this to standing waves. Consider a diving pool that is \(5.0 \mathrm{m}\) decp and \(10.0 \mathrm{m}\) wide. Standing water waves can set up across the width of the pool. Because water sloshes up and down at the sides of the pool, the boundary conditions require antinodes at \(x=0\) and \(x=L\). Thus a standing water wave resembles a standing sound wave in an open-open tube. a. What are the wavelengths of the first three standing-wave modes for water in the pool? Do they satisfy the condition for being deep-water waves? Draw a graph of each. b. What are the wave speeds for each of these waves? c. Derive a general expression for the frequencies \(f_{m}\) of the possible standing waves. Your expression should be in terms of \(m, g,\) and \(L\) d. What are the oscillation periods of the first three standing wave modes?

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