QUIZ 3 CH. 15, 16
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| Exercise 15.4: Ultrasound Imaging; Exercise 15.6; Exercise 15.12: Speed of Propagation vs. Particle Speed; Exercise 15.16; Exercise 15.20; Exercise 15.24; Exercise 15.38; Exercise 16.33; Exercise 16.35; Problem 16.70 ; |
| The traveling and standing wave problems are straightforward and will be considered last; let's first explore interference: CH. 16; we assume a common detector receives waves from two sources and the path difference between the two waves produces constructive or destructive interference. Remember constructive means path difference = n*lambda and destructive means path difference = (2*n + 1)*(lambda/2); n = 0, 1, 2, 3,4, ..... Usually the path difference is found using 1D or 2D Euclidian geometry as shown below. Meanwhile, the sample exams' treatment of interference here defines/extends the envelope. |
| 33. In this case, the path difference is fixed at
2.00 m but we are changing the wavelength to achieve conditions of constructive
or destructive interference. (a) constructive: 2.00 = n*lambda, n = 0, 1, 2, 3, 4, ..But wave speed = v = w/k = lambda*f. Let v = 344 m/s, the speed of sound in air at room temperature 20 degrees Celsius. Thus 2.00 m = n*v/f, or f = n*v/(2.00 m). What is the lowest non -zero value of f? (b) destructive: 2.00 = (2n+ 1)*(lambda/2), n = 0, 1, 2, 3, 4, with wave speed = v = lambda*f. Thus 2.00 m = (n+1/2)*v/f, or f = (n+1/2)*v/(2.00 m). What is the lowest non-zero value of f? |
| 35. LET DA be the distance between you and
speaker A and DB how far you are from speaker B. Thus DA +
DB = 12 m. DB - DA = n*v/f for constructive interference, n = 0, + or -1 + or - 2, + or -3, ..... and DB - DA = (n + 1/2)n*v/f for destructive interference, n = 0, + or -1 + or - 2, + or -3, ... Suppose you are at a position of constructive interference; then DB - DA = n*v/f for a given n. If you now decrease DB - DA, you would be walking toward speaker B. Let's look at this problem more analytically then check with our physical intuition: First let's solve for DB : For constructive interference DB + DA = 12 and DB - DA = n*v/f,
which leads to DB = n*v/(2f) + 6. Suppose we have constructive interference; thus DB - DA = n*v/f = n*lambda and DB = n*v/(2f) + 6. As we walk toward B to cause destructive interference we decrease DB from n*v/(2f) + 6 to (n - 1/2)*v/(2f) + 6. We see the difference is -v/(4f) = -lambda/4, so you'd walk one-fourth a wavelength toward B. Note if DB deceases by lambda/4, DA increases by lambda/4 and the path difference DB - DA = lambda/2 as you would expect for destructive interference. |
| 70. DISCUSSIONS SOON. |