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Safari File Edit View History Bookmarks Window Help edugen.wileyplus.com FIU Logirn (43) saadsalmeen96@gmail.com - Gmail wileyplus Yahoo Search Results Yahoo Search... WileyPLUS WileyPLUS: MyWileyPLUSI Help I Contact Us I Log Out Engl besity WileyPLUS Halliday, Fundamentals of Physics, 10e Calculus-based Physics I ⅈ (PH 201-202) Home Read, Study & Practice Assignment Gradebook ORION Downloadable eTextbook Assignment> Open Assignment ON BACK NEXT ASSIGNMENT RESOURCES 201-22 Chapter 17, Problem 013 A sound wave of the form s = sm cos(kx-ut + ) travels at 344 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.03 m is at its 006 hapter 17, Probl 013 Eng ximum positive displacement of 6.00 nm and air molecule B at x = 2.10 m is at a positive displacement of 2.00 nm. All the molecules between A and B are 020 030 040 intermediate displacements. What is the frequency of the wave? Number the tolerance is +/-296 Units Engli per LINK TO TEXT LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE Review Results by Stu En Question Attempts: 0 of 6 used SAVE FOR LATER SUBMIT ANSWER pire

Explanation / Answer

speed v = 344 m/s

particle A is at maximum displacement so speed for A = 0

6*10^-9 = sm*cos(k*2.03 - wt + phi )

speed v = ds/dt = 0

-sm*sin(kx - wt + phi) = 0

0 = -sm*sin(k*2.03 + phi)

k*2.03 - wt + phi = 0

- wt + phi = -k*2.03

for particle B

2*10^-9 = sm*cos(k*2.1 - wt + phi)

2*10^-9 = 6*10^-9*cos(k*2.1 - k*2.03)

cos(k*2.1 - k*2.03) = 1/3

k*0.07 = 1.23

k = 17.6

frequency f = w/(2pi)

speed v = w/k=====> w = v*k

f = v*k/(2pi)

f = 344*17.6/(2*pi)

f = 963.6 Hz

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