t1fall02sol - PH‘ISHJ 9.301: Fe.” 2902. 715i"....

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Unformatted text preview: PH‘ISHJ 9.301: Fe.” 2902. 715i". 3'45! - Salu‘iflvny 1. Two sinusoidal traveling waves are moving in the positive x direction. They have the same amplitude, call it A, and the same speed, frequency, and wavelength; but there is a relative phase between the two. If the superposition of these two disturbances results in a net disturbance whose amplitude is also A, thentheirrelativephaseindegrees is: xii“ : AS”; (&yfwt)_+fl 1M {Axrwf Ho) (1) 30 (2) 45 (3) 60 (4) 90 @323 (6) 180 “1”” 3 f1 0 CM (‘91) ‘5?“ U»! f wt + W») A :3? (05 = y; 1-) $41L00$p3l200 2. Two triangular-shaped wave pulses (isosceles triangle) approach each other from opposite directions. 5:: They each have a speed of 5 111’s relative to the ground, their individual Widths are 0.4 in each (base of the triangle), and one pulse is inverted relative to the other. How long after their peaks are 2 m apart will the net disturbance be as shown in the lower of the two pictures? [see pictures below] In apart at t = sec 1 0.20 seconds @024 seconds 3) 0.40 seconds (4) 0.48 seconds {5) 0.60 seconds (6) 0.80 seconds TI“ hk , a)” m nurse Macs '- f" 2.21;; , 3. The equation of a transverse wave propagating a orig—aflor ontal string with the disturbance in the y direction is given by y = 0.04-cos(20a-t + 81px — its?) . The units of time are seconds and all length units are in meters. The wavelength for this disturbance is: 1 0.10m gfizl 2 .25m A 0.50m )3 3115’” (4) 8a in .__.—-—_ (5) 4n ni (6) an? m 4. For the wave of problem #3, the speed and direction of propagation of the disturbance is: (1) 0.4 mi’s , positive it we}, haw; x d H“ { H'H (2) 0.8 this , negative x w — - amt - i3i l.25mts,positivex bl ‘ “ 6," ~ 2.5 mfg 2.5 m-ls , negative x _ 5) rtf2 rrv’s , positive y (6) 47: mi s , negative y a. For the wave of problem #3 the acceleration of the piece of string in rat-"s2 at t = 2 s and X = (1.348) In is: (1) 0 -' . . (2) 1.3 Y z “"97 3 -' ( “1731039) 60501011 + g .. fig g M Hoax .05!an cos (- Il‘g)= _7q mjsx 5 158 H ._ .ho” is; 316 M“ W 6. A string of length 1 m and mass 0.01 kg is clamped at both ends. If the second harmonic resonant frequency is 50 Hz, then the tension in the string is: awhum, w 2- .. ,3. (1) £31,131 N val/7. — xc-ISO m/s L1. ‘ ZON ‘ ‘v‘ MK 9.5.0? “3' {)Tflxzs‘oofzs-N (5) 10N __-:.=-' ______. (6) SN 7. A spherical sound wave is emitted by a point source. The intensity of the sound 50 m from the source is 10'8 me2. Assuming no loss of power as the wave propagates outward fmm 1he source, the intensity of the sound 25 in from the source, in units of 10'8 Wr’rnz, is about: (1) 0.06 I" or 9,.» al- lax]; le-i Ankm Rims. fhle‘me (2) 0.25 (3) 0.50 2.00 4.00 [6 16.00 8. For the source described in problem #7; relative to the sound level 50 In from the source, the sound level 200 m from the source is about: [Find the sound level at 200 In minus the level at 50 n1.] Tug ' I" - I v (1) --4dB JSM— “35.? '“—' '04:; “Ways—V =_- I01? «.3" (2) --6 dB _ i l o ’“s v (3+1de AAB : {01353 (L) : “05?”, r __ z {4) —4 as w———___. ‘t l “I 8 ' —6 dB 6) —12 dB 9. Two speakers that are positioned as shown below, are driven in phase by a common source. What frequency of sound waves will give fully constructive interference at the point P in the figure below? (1) so Hz (2) 255 Hz 340Hz (4) 425m. (5) 4s0nz (6) 7651-12 10. organ pipe is closed at one end and open at the other end. If a standing wave disturbance is create with three intermediate nodes (not counting the one at the closed end) and the length of the pipe is 3.5 m, then the resonant frequency is: 1 ‘“ "Jfixr 3.5m X :iXBA'm: 2.9m 1r 170Hz (4)194Hz {1: 7f ’= 3'40 :170H_ (5) 255 Hz ____. 2.0 5 (6) 290 Hz “a z 11. Two police vehicles are closing in on a stationary fugitive from opposite directions. Each vehicle is moving at 8.5 Ms. If the frequency of each squad car’s siren is 100 Hz, then the beat frequency heard by either of the two drivers is about: "OHM/2Q 3-40 is; 4-- :: - ' foo S'- El) 5H2 It“? (1r—-'L’})£ (suggests PM“ ?.5Hz 3) 10Hz ‘ ' (4) 15Hz l‘i“ 2‘ ‘ l+(‘/1<thoo (5) 20Hz 1-00;) (6) 110Hz I} o w m : l3.th — R, e U“: n'lw [IF-s 12. The force on the charge +Q located at the origin due to the other three charges, that are shown below, is toward: positive it and positive y 2) positive x and negative 3! (3) negative x and positive y (4) negative X and negative 3! (5) no direction since force zero 13. Suppose the charge at the origin inproblem #12 is relocated to infinity. The magnitude of the electric field at the origin due to the remaining three charges that are shown in problem #12 is about: ‘1 I? ‘= M 0.2%? '3 * 1 E- (a s ’3'“ ‘43 (3)0'71‘6Ql32' IE'l :- IQSL straw-‘2' .2 oat-m FLO/a1 u1 ' (4) 1.41<..Qra2 . (5) Linorag . refit—fie .~——-——-—-—r—'____.__———-—— (6) 2.4mm2 . 14. Two rings of charge, each of radius a, one with total charge +Q centered on the x—axis at x = —a, the other with total charge —2Q centered on the x—axis at x = +a, have their planes perpendicular to the x-axis. The magnitude of the electric field at the origin is about: [ See figure inprohlem #15 ] (1) 0.18nQra2. —=- Mg 0_ A g L 4 E" " la)" + film * (2) 0.411eQra2. — ~——-——- 1 omega? A (ti-E)“ l“: “l a 1.1keQr'a2. E 3 LL 2- (5) 1.5keQr'a2. 1. (6) llwaz. fir/4L r 15. Consider the situation described in problem #14 and shown here. The potential at the origin is about: «that ran (1) +2.1keQr'a f- (2) +1-0keQe’a 3 +0.71k.Qra a: - ABgGP. : .0 19733 (4)} —0.71k.Qia Y; on X (5) ‘LOl‘tea V cc - 0.7I U (6) —2.1k.Qra a, 16. Consider two semicircular rings of charge connected by two lines of charge as shown below. The charge is uniformly distributed on each of the rings and on each of the lines; but may be different on the two rings and on the lines. The semicircle of radius R carries charge —.Q while the semicircle of radius 2R carries charge +2Q. Q is a positive number for these considerations. The y—corrrponent of the electric field at the common center of the arcs, if x is positive to the right and y is positive upward is: (1] +2ch,t':rtR2_ hurl-I11 tin X " Lumfgd‘nt “mod, (2) +k5Qfl’lTR2 v Y “5? “fig “minis 1mm (3) 0 "7 5 v. (i) X (4) «east. Ev _ redial/a . (5 —2kle:IrR2. 4 . «mimosa? 11: '28 -2Q . 1?. For the charge distribution shown in problem #16, the potential at the common center of the arcs is: (1) Exactly +keQr’R Th Comhrmal pa “hi-Id (2) Exactly —l(leR _ . ‘ (3) Exactly —2kBQ.«'R +E‘? +106 SQMHH‘CICJ (“1” (4)0 Cufiai tuck OHM’P'.TL{ @2:223:12ZIEC'EE‘biil‘fiflifi“(2)“(3) puma PM we M am; It hm cm be prams/Lat not“);th‘V FOR PROBLEMS 18, 19, 20 CONSIDER THE FIGURE SHOWN HERE. In all these problems, Q is a +w positive number, and r is the radial distance from the center of the concentric spheres. The-re are two R conducting, concentric, spherical, shells; the one between R and 2R has a total charge —3Q , and the one f‘M between BR and 4R has a total charge +2Q. The system is in static equilibrium; and the potential infinitely far from the common center is zero. -3Q conducting 18. The radial electric field in the region 2R < r < 3R is: [positive is radial outward, negative is inward] (i) _ 02 PAW. : —393/€u (—l hQr’r (3) more E .. -§9L_- '2': @141 5 ,+2k¢er3 are .3: @319th 19. The charge on the surface at r = 4R is: iii) 0 Nasal +3014: 1" =3R 9,, im 5:9 a; wvavath -Q .. (3) ‘2Q find“ HQ" 3mm “10 3) 949a” " (P (4) +3Q (5) +2Q {6) +Q 20, The potential at r 3 3R, if the potential at infinity is zero, is: (1) —3k,QxR ‘ _ 93W 7- ‘9 Xe “(bub any'ywy Var) ‘ ‘ (2) r3keQ-QR Y— ? $3313 wt r : HR Vim) : 49/“ Q i323??? 3”” W3“): VH3) {mariachi Venn-MM F: ...
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This note was uploaded on 03/19/2008 for the course PHYS 2306 taught by Professor Ykim during the Fall '06 term at Virginia Tech.

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t1fall02sol - PH‘ISHJ 9.301: Fe.” 2902. 715i"....

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