HW3-4Sol

HW3-4Sol - Oregon State University Physics 202 Winter 2010...

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Unformatted text preview: Oregon State University Physics 202 , Winter 2010 HW3-4 3 (due Feb. 1 at 5:00 p.m.) 4 Page 1 Oregon State University Physics 202 Winter Term, 2010 HW3-4 Solutions 1. Chapter 15, Problem 62 (page 505). a. The wave is traveling in the –x direction (indicated by the + sign in the cosine argument). b. Looking at the two parts of the argument of the cosine function: The wavelength: λ = 2.40 m. The period: T = 0.200 s. The frequency: f = 1/ T = 5.00 Hz. T T The wave speed: v = f λ = 12.0 m/s c. y (.20 m, 0.50 s) = (3.0 cm)cos{2 π [(.20/2.4) + (0.50/.20)] rad} = –2.60 cm 2. Two sources of sound are located on the x-axis, and each emits power uniformly in all directions. There are no reﬂ ections. One source is positioned at the origin. The other source is at x = 132 m. The source at the origin emits four times as much power as the other source. Find the two positions along the x axis where the two sounds are equal in intensity. A B 0 132 x First note: Since one speaker is more powerful than the other, in order for the intensities of their sounds to be equal to you, you will need to stand somewhat nearer to the weaker speaker (B). There are two places that could be: somewhere between the two speakers (but nearer to B); and somewhere to the right of speaker B. (Note that there is no way you could ever hear equal intensities by standing left of speaker A, since it—the more powerful—would always be nearer to you there.) As it turns out, if you set up the problem to solve for one position (either one), careful algebra—i.e. using both solutions of a square root or quadratic—will give you both solutions. So, suppose you stand somewhere in between the speakers, at a distance X from the origin. That means you are standing at a distance X from speaker A and a distance of (132 – X ) from speaker B. If X is the right spot, the intensities are equal: I A = I B In other words: P A /(4 π r A 2 ) = P B /(4 π r B 2 ) But weʼre told that P A = 4 P B . Therefore: 4 P B /(4 π r A 2 ) = P B /(4 π r B 2 ) And weʼve observed that r A = X , and r B = 132 – X: 4 P B /(4 π X 2 ) = P B /[4 π (132 – X ) 2 ] Simplify, by dividing both sides by P B /(4 π ): 4/ X 2 = 1/(132 – X ) 2 Now, at this point, you have a choice. You can either clear the fractions on both sides—to create and solve a quadratic equation (perfectly logical and correct); or you can do this: Multiply both sides by X 2 : 4 = X 2 /(132 – X ) 2 Take the square root of both sides: ±2 = X / (132 – X )...
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This note was uploaded on 03/27/2010 for the course PH 202 taught by Professor Staff during the Spring '08 term at Oregon State.

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HW3-4Sol - Oregon State University Physics 202 Winter 2010...

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