/*! 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 27 Energy Output. By measurement yo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 27

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is \(0.026 \mathrm{~W} / \mathrm{m}^{2}\) at a distance of \(4.3 \mathrm{~m}\) from the source. (a) What is the intensity at a distance of \(3.1 \mathrm{~m}\) from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

Short Answer

Expert verified
a) The intensity at a distance of 3.1 m from the source is 0.048 \, \mathrm{W} / \mathrm{m}^{2}. b) The sound energy emitted by the source in one hour is 128332.33 J.

Step by step solution

01

Apply the Inverse Square Law

To find the new intensity at a distance of 3.1 m, apply the inverse square law. This law states that the intensity of a wave is inversely proportional to the square of the distance from the source, i.e., \(I_2 = I_1 * (d_1 / d_2)^2\), where \(I_1\) is the initial intensity, \(d_1\) the initial distance, \(I_2\) the final intensity, and \(d_2\) the final distance. Substituting the given values, \(I_2 = 0.026 * (4.3 / 3.1)^2\).
02

Calculate the New Intensity

Now, perform the calculation for \(I_2\). So, \(I_2 = 0.026 * (4.3 / 3.1)^2 = 0.048 \, \mathrm{W} / \mathrm{m}^{2}\).
03

Apply Energy Time Relation

For the second part, the energy emitted by the source in one hour is calculated by using the formula for power \(P = E / t\), where \(P\) is the power, \(E\) the total energy, and \(t\) the time period. Given that power output remains constant, the power is equal to the initial intensity times the area of the sphere at the initial distance \(P = I_1 * 4 * \pi * d_1^2 = 0.026 * 4 * \pi * 4.3^2\). As power is energy per unit time, and we are asked to find the energy in one hour, we can find the energy by multiplying the power by the time in seconds \(E = P * t = P * 3600\).
04

Calculate the Total Energy

Finally, substitute the values into the formula and calculate the total energy. So, \(E = P * 3600 = 0.026 * 4 * \pi * 4.3^2 * 3600 = 128332.33 \, \mathrm{J}\).

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!

Key Concepts

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

Sound Waves
Sound waves are vibrations that travel through the air or another medium and can be heard when they reach a person's or animal's ear. When an object vibrates, it causes the surrounding particles to vibrate as well, which, in turn, causes the adjacent particles to vibrate. This chain effect carries the sound wave through the medium.

Sound waves are longitudinal waves, meaning the vibrations occur in the same direction as the wave is moving. These waves are characterized by their frequency and wavelength, which determine the pitch and tone of the sound. In a three-dimensional space, such as when emanating from a point source, sound waves radiate outwards in an expanding sphere.
Power Output Calculation
The power output of a sound source can be determined by examining the intensity of the sound waves it emits. Intensity refers to the power per unit area carried by a wave, and is measured in watts per square meter (\(\mathrm{W/m^2}\)).

To calculate the power output (\(P\)), we can use the initial intensity (\(I_1\)) and the surface area over which the sound wave spreads at a certain distance from the source. For a point source emitting sound equally in all directions, this surface area is the surface of a sphere with a radius equal to that distance: \(A = 4\pi r^2\). Hence, the power output formula is:\[P = I_1 脙聴 A = I_1 脙聴 4\pi r^2\], where\(I_1\) is the intensity at the radius \(r\).
Energy-Time Relation
The energy-time relation in sound physics is instrumental for calculating the amount of energy emitted by a source over a period of time. Power (\(P\)) is defined as the rate at which energy (\(E\)) is emitted or converted per unit of time (\(t\)). Thus, the formula for power in terms of energy and time is given by:\[P = \frac{E}{t}\].

Consequently, if we want to find the total energy emitted over a specific time period, we can rearrange the formula to find energy: \[E = P \times t\]. By knowing the power output of a source and the duration for which it has been emitting energy, we can calculate the total energy output during that time.
Inverse Square Law Application
The inverse square law is fundamental in understanding how the intensity of sound waves changes with distance from its source. It states that the intensity (\(I\)) of a sound wave is inversely proportional to the square of the distance (\(d\)) from the source, mathematically expressed as \[I \propto \frac{1}{d^2}\].

When applying this law to calculate a new intensity (\(I_2\)) at a different distance (\(d_2\)), we use the following formula:\[I_2 = I_1 \times \left(\frac{d_1}{d_2}\right)^2\], where \(I_1\) is the original intensity at distance \(d_1\). In essence, the inverse square law illustrates that as you move away from a sound source, the intensity decreases quickly; specifically, if you double the distance, the intensity becomes one-fourth.

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

Standing waves are produced on a string that is held fixed at both ends. The tension in the string is kept constant. (a) For the second overtone standing wave the node-to-node distance is \(8.00 \mathrm{~cm} .\) What is the length of the string? (b) What is the node-to-node distance for the fourth harmonic standing wave?

Scale length is the length of the part of a guitar string that is free to vibrate. A standard value of scale length for an acoustic guitar is 25.5 in. The frequency of the fundamental standing wave on a string is determined by the string's scale length, tension, and linear mass density. The standard frequencies \(f\) to which the strings of a sixstring guitar are tuned are given in the table: $$ \begin{array}{l|llllll} \text { String } & \text { E2 } & \text { A2 } & \text { D3 } & \text { G3 } & \text { B3 } & \text { E4 } \\ \hline f(\mathbf{H z}) & 82.4 & 110.0 & 146.8 & 196.0 & 246.9 & 329.6 \end{array} $$ Assume that a typical value of the tension of a guitar string is \(78.0 \mathrm{~N}\) (although tension varies somewhat for different strings). (a) Calculate the linear mass density \(\mu\) (in \(\mathrm{g} / \mathrm{cm}\) ) for the \(\mathrm{E} 2, \mathrm{G} 3\), and \(\mathrm{E} 4\) strings. (b) Just before your band is going to perform, your G3 string breaks. The only replacement string you have is an E2. If your strings have the linear mass densities calculated in part (a), what must be the tension in the replacement string to bring its fundamental frequency to the G3 value of \(196.0 \mathrm{~Hz}\) ?

A deep-sea diver is suspended beneath the surface of Loch Ness by a \(100-\mathrm{m}\) -long cable that is attached to a boat on the surface (Fig. \(\mathbf{P} 15.77\) ). The diver and his suit have a total mass of \(120 \mathrm{~kg}\) and a volume of \(0.0800 \mathrm{~m}^{3} .\) The cable has a diameter of \(2.00 \mathrm{~cm}\) and a linear mass density of \(\mu=1.10 \mathrm{~kg} / \mathrm{m} .\) The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat. (a) What is the tension in the cable at its lower end, where it is attached to the diver? Do not forget to include the buoyant force that the water (density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) ) exerts on him. (b) Calculate the tension in the cable a distance \(x\) above the diver. In your calculation, include the buoyant force on the cable. (c) The speed of transverse waves on the cable is given by \(v=\sqrt{F / \mu}\) [Eq. (15.14)]. The speed therefore varies along the cable, since the tension is not constant. (This expression ignores the damping force that the water exerts on the moving cable.) Integrate to find the time required for the first signal to reach the surface.

You are designing a two-string instrument with metal strings \(35.0 \mathrm{~cm}\) long, as shown in Fig. \(\mathrm{P} 15.52 .\) Both strings are under the same tension. String \(S_{1}\) has a mass of \(8.00 \mathrm{~g}\) and produces the note middle \(\mathrm{C}\) (frequency \(262 \mathrm{~Hz}\) ) in its fundamental mode. (a) What should be the tension in the string? (b) What should be the mass of string \(S_{2}\) so that it will produce A-sharp (frequency \(466 \mathrm{~Hz}\) ) as its fundamental? (c) To extend the range of your instrument, you include a fret located just under the strings but not normally touching them. How far from the upper end should you put this fret so that when you press \(S_{1}\) tightly against it, this string will produce \(\mathrm{C}\) -sharp (frequency \(277 \mathrm{~Hz}\) ) in its fundamental? That is, what is \(x\) in the figure? (d) If you press \(S_{2}\) against the fret, what frequency of sound will it produce in its fundamental?

A string or rope will break apart if it is placed under too much tensile stress [see Eq. (11.8)]. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has density \(7800 \mathrm{~kg} / \mathrm{m}^{3}\) and will break if the tensile stress exceeds \(7.0 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\). You want to make a guitar string from \(4.0 \mathrm{~g}\) of this type of steel. In use, the guitar string must be able to withstand a tension of \(900 \mathrm{~N}\) without breaking. Your job is to determine (a) the maximum length and minimum radius the string can have; (b) the highest possible fundamental frequency of standing waves on this string, if the entire length of the string is free to vibrate.
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Work Energy and Power

Read Explanation

Electrostatics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physics of Motion

Read Explanation

Medical Physics

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.