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9_pdfsam_Phy63S10-Q1-4-Solutions

9_pdfsam_Phy63S10-Q1-4-Solutions - P113763 Spring 2010 1...

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Unformatted text preview: P113763 Spring 2010 1,! = (02an k=2nl7t VZM \J (A. : 5‘- Total phase- or path—difference conditions for interference (in units ofl): (l) 0, 7L, 27L, 37L, (2) M2, 31/2, 51,12,712, For standing "waves: (3) 1/2, 3M2, 51/2, 712, (4) M4, 3M4, 5M4, 7M4, Normal modes: (1) n0], /2) = L ; (2) (2n+1) 7L“ /4 = L (n an integer) 1 = again? 10 10g (1/1,) ' f’ = (v +5 V1)/(v — v;) r 2Acos[21n|f1—f2|t/2] 1. A one meter long guitar string, held under tension, is fixed at the two ends (cannot move at the ends). A guitarist plucks it in the middle to make a note. The excited normal modes are predominantly even harmonics of the fundamental (1 point): . o 2. Two Doppler shifted frequencies f1 and f2 are compared. With both source and detector stationary, the frequency of the sound is f. The frequency fl is measured when the source is stationary and the detector is on a car approaching the source at V6. Then, .152 is measured when the detector is stationary and the source is on the 'Car moving away from the detector. The speed of sound is lOvc. It is found that (2 points): ®r1>r2, and f1=1.1f, f2=f/1.1 b. f1>f2,and f1=f/0.9,f2:r/1.1 c. f1<f2,and f1=1.1f, f2=f/0.9 a. f1<f2,and f1=f/1.1, f2=f/0.9 e. flzfz, and $20.9 f, f2=0.9f. ...
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