Solutions_p125MidW10

Solutions_p125MidW10 - UNIVERSITY OF WATERLOO DEPARTMENT OF...

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Unformatted text preview: UNIVERSITY OF WATERLOO DEPARTMENT OF PHYSICS PHYSICS 125 WINTER 2010 MIDTERM EXAMINATION Date: February 24, 2010 Time: 7:00-9:00 pm. NAME (Print): I.D. #: PROFESSOR: SECTION #: SIGNATURE: IMPORTANT INSTRUCTIONS - READ CAREFULLY TO AVOID PENALTY 1. Use PENCIL. Mark your ID# and three digit “Section number”. Check that you have not missed, or put double marks in any row in the answer field. (Keep sheet face down when not in use.) 2. Fill in name, etc., at upper right on computer sheet and on this page. 3. HAND IN SHEET AND BOOKLET SEPARATELY. Booklets will be returned to the students later. 4. Check that your exam has 13 questions. All questions are of equal value, so don’t get hung up and miss the easy ones. Formula sheet should be attached 5. There is no penalty marking for incorrect answers. However, in the event that the computer sheet cannot be completely read, credit will be given to correct answers only if work is shown in the space below the questions in this booklet. 6. If you need more space for your calculations, use the blank pages at the end of the booklet or use the back of the previous page and label it with the question number. 7. Good luck!! g = 9.3 111/52, 10 =10"2W/m2. Circle your Professor: R. Jayasundera (Mech) R. Jayasundera (Management) Physics 125 Midterm Examination Winter 2010 Page 2 of 10 l. A particle executes simple harmonic motion with amplitude 5 cm and frequency 12 Hz. At time t = 0 it has a positive displacement and is moving in the positive direction at 2 m/s. When will this particle have the same velocity next? Answer in ms (1 ms = 10'Jsec.). (b) 68 (fa) 82) (d) 98 (e) 44 L7; ,( ,1 fl” bx/U if 05 m wrwflfiyfl/‘i‘ a; 7{(.0J 7C) (Zr/1‘ l/(O) - ’0 7 7 I flflrmyj 7“ ’ "/4th é? fl ‘ n4, (‘Pj3‘2 ""7/ g/‘6 / u/ ”' L , _ ¢_ 7'; (ax/014v“) A L¢</\( '1’} ‘ 0’)» ' S/‘/1( 6 ’ 0" {>77 KM" "v/ 2]": , fl finer: #«t/ay, [u y - , .. 5/ 7/ =— 5- ' 7 £‘\‘( 9 ‘ 1' 3 9‘” 'U 4 ; ’7‘4’Ck(/t f)!” 2(a’4'5'jl‘!‘ A?“ a ; «'2.- K r” 7‘“ fiM : 5‘6 M5 1%; w *‘ M. 7 g: / ,1 /7 1/ /q N541. fl" 6 9— ,é‘+5.'7) at can - O > O 0)- (‘34 ( 294/ ‘ W021,an 5'-7 : W», 5/2, W4 6"" / é A1421wa 2%!" “"7 "'57" / H 7“ 6 2 0 41( 2. A wire 20 m long and having a mass of 200 grams is kept taut, under tension of 50 N. A pulse is generated at one end of the string and 50 ms later a second pulse is generated at the other end of the string. Where will the pulses first meet (find distance from the end the first pulse was generated)? Answer in m. 575 (3)102 (c) 12.7 (d)14.4 (e)15.3 A 9.0 m 50 m) (an W4C 54-.»A'y'cc/é" flu" (¢,(' [1) ,4/5 : 3.5“ Sec/Y7 —.¢=—-"’-" rim/J Qwoflam «.m‘// 5/44/4 :L/‘ C QM"! 3/410 72“?” [tau-c SC’C’PW 5%kwo5 '77u7 oar/K We f /1L>L€fl//L<Jr‘7 [u #‘W‘l. ext. at! D M): neg-1H": 2J3 flD :» M73 : IL? (/7 Physics 125 Midterm Examination Winter 2010 Page 3 of 10 3. A simple pendulum consists of a ball of mass m on the end of a /\ string of length 8 = 1.2 in. There is a peg P a distance I /2 below D the suspension point as shown in the diagram. Find the period for é small oscillations. Answer in seconds. /, ’ (a) 1.55 w (c) 2.18 (d) 3.14 (e) none of th / , 4 C] . 5 ll] z” , / 2 t / / - a l/ [57/ J fl: m v e ‘ c,.gr / x i / 5' 5’ K. ,— a / 2 / 67 0 7g / “A I l/ : / O / 9- ' 9 S . 2 ’ / ‘ ( < 4. A uniform meter stick of mass m is pivoted at one end and oscillates in simple harmonic xx motion with period T see. A point mass (small mud ball) is now stuck at a point “ q ”, , which is “ y” meters below the pivot point. If the mass of the mud ball is the same as the Z/ mass of the meter stick and the new period of oscillation is also T sec, (same as before) / how far below the pivot point is the mud ball. Answer in m. (a) 0.25 (b) 0.33 (c) 0.5 (e) 0.75 /, 7N7 J: Pivot /’ flay/7 '2—/ 1m (Z / / (L L — 1/71 f l rat/a — é ”’ "’77 3 7’" .- my rgz‘) 3 7C M/QZé/i: lm/5(‘H5ZLJ ’fiy / ‘ Alvj(,f2»() Physics 125 Midterm Examination Winter 2010 Page 4 of 10 5. A uniform hoop of Radius R and mass M is free to rotate about an axis through its cm. Two springs are connected to the outer edges and a torsional spring at the ,_/f\ centre. If the system is set in simple harmonic motion, '/ a which statement describes the period of small angle 1 / 7 7 " oscillation? (The moment of inertia of the hoop is MR2 / and the spokes are massless.) M MgR (a) 2’\'}1z(k1+kz)+x (b)1\x)R(k,—k2)—K ’ ‘ fk +)< )- 7/14/5129 .A/LQ : ,- [k‘.+lz1+)c/flz (3 (It; m A / m JV — «111x o W , In a ’2 fl; / l M x A. [a +)(/‘ a __ a .r_ .- . Cal: (1' L K (“o/a}~2’l/ b,/-1//' k M 7 “Wk” [4 6. Astring tied toasinusoidal oscillator atPand running over 9 "'5- ~\. asupport at Q, is kept taut byablock massm. The Q - ) separation between P and Q is 1.2m. An oscillator is a éy// producing a frequency of 120 Hz. P and Q can be Oscillfl‘l’d thought of as nodes. A standing wave pattern will be m produced when a mass of 286.1 g or 447.0 g, but not for any intermediate mass. Find the linear density of the string. Answer in g/m. (a) 0.54 (b) 0.68 (c) 0.72 (e) none of these F w a rml':9~Eé-/ f f [77/( 77:6415 w")? 2. Wires? 77 WA - m m. , ’7)? ' -————~ 'L‘ 7’4? / Z 3 f) %‘ a W ‘ m m/ c A—A-Tg J I : ’ I” I, ' ’- 0’ m .2fl_../«' W 1/)- — 7L; : 2 23‘ :‘7 '77: Firm M ’"9/ am '/ 0’) km I6 114. /"l i 7”: I 7” 3/ LY “gm” ”°/ H“’5'. H-w—éu~ W '1‘, ’ m: S- 3,. : 31—— m:/lu : (9";9 .r/t- 2:: Physics 125 Midterm Examination Winter 2010 Page 5 of10 .13 7. Two identical speakers emit identical sound waves of frequency 9 :0 A 170 Hz uniformly in all directions in space. Each speaker has a ‘ power of 1 mW (10" W). A point P is 2.00 m from one speaker [/7 ’ and 3.00 m from the other. If the speakers are driven in phase and the speed of sound is 340 m/s, find the loudness at point J 1’ “p” when both speakers are on. Answer in dB. 31 IO (b) 73 (c) 84 (d) 92 (e) none of the above Q : 39/(272/‘4 A’frlm . (37’: 9/L_ :9 d/Cé/WM‘A“ no /’r r/ - 4’, i .51 t ' ) L . j‘l/lu (“67HMj1roé’ (4 ” fl/‘é A 5/ ) L (fl I . ,v A 7 - “~41 we ex— A: ._ ,4 L 3,; :3 fl/ ‘ ' '74“ fl/=/.5fl /‘ i f .1 H I - /‘ L. flI/fl —(fl}'fl/) : 0.5‘fl . , j,fk,f.f (Glam/q} T 0_ 15. 2 I 7 w ' - 0.34 1 ' I 1 ___ .» ' A : ,wwML. .. 493%; 1,1 5 [277),]. ’ 1/ 77(7) ) ' (//3 / é , 1.. /j ' ' ire-Jr” Q 501/0 V”/m /-—_ 8. The sound level from an exploding volcano is measured at two points, P. which is at n f D meters and P2 which is at r; meters. These points are along the same line from the centre (4 I of the volcano. The difference in sound levels between P1 and P2 was found to be 20 dB. " If the two points are 2 km apart find the values for n and r2. Answers are in km. (b) 033,233 (c) 0.44, 2.44 (d) 055,255 (e) none ofthese h. Va\c..“° Pr D '_ : IC) Q0 ;/0 1/6- ZJ] 777/ [1. f 7/ 6/2 , I0 / a“;l:/O~ rylfl’f‘l: 1 z} /- - rj/ ’ 2‘ muu 1" I 9/77,- "' 1', ' (7 )1 M 22 s 7’; :1 2am III/7‘2 _/ Physics 125 Midterm Examination Winter 2010 9. 10. Page 6 of 10 When a tuning fork of frequency 500 Hz is held above the tube as shown, resonances are found when the water level is at distances L = 16.0, 50.5, 85.0 and 119.5 cm from the top of the tube. What is the speed of sound in the air? Answer in m/s / (a) 335 ‘ (c) 355 (d) 365 (e) none of thes ’ .V w Y ’(‘D ’ (M Ln 9‘ ~ a : 69ch W C 67'“ U” 0'69 HUD " 375 '"/5. If the total energy of a harmonic oscillator is reduced by 1/3, what is the change in the amplitude of the oscillations? m3 1‘» (6>"“J§“” (c) 1/3 (d) 3 ’L g, = j /e a, M 52-7 /4 5' ‘ £1 £12 _} 21' fit 1- L- z 1.. A; ' 3’9’ INJ. fl?- Aaa Nadaaoé‘ 3 Physics 125 Midterm Examination Winter 2010 Page 7 of 10 Reminder: A tuning fork vibrating at 512 Hz falls from rest and accelerates downward at 9.80 m/ s2 . How far below the point of release is the tuning fork when sound having a frequency of 485 Hz reaches the point of release? Take the speed of sound in air to be 340 m/s. Answer in m. v=vo+at (y‘Yo)=vot+%at2 v2-v§=2a(y—y°) , V x) (a) 18.2 (c) 20.1 (d)21.2 (e) none ofthese f3 0 l 9 fl f r f I 1 (2L~L-( Sacra 4»; f. "'2" f y m (/9 1 A) (ML-y '3 l‘ O l t _- I if?) ¢ 573‘ (fiétg—Or-D) 3) 0' ’9 390 f”)- : ~ b '2}, .‘2 [3 C7 l: '7txfl/N] (fry/L 5/6“ A (4:14" W7/ 89 (2)1 Q( / f)ar \ ‘ 2. L f 3,‘ ‘1 ( ry‘, J :7 / 1/1‘ \\. U " %_>_ _ HM’ _ g z /lq4 J \ W ,(1/ 1 l5’ 3 In) 7704) / ‘ baud“ __,,_—-—v /"'/< . ‘ (7L2.'/ M A, W .Mun S ' /7£W4/g Dioyyfauv [4(4Hg'flcwdém f: “M 12. / (a) (b) (C) flu. (Ciro/s (7’ \ Cc ’ XL‘W, C V4447 C( . , 1. (/9 anew)" r imam-u l . 714,644,425 a l Vt] .1 [Joy/n Clo, LC( g) ‘5 f (.0? __ I; 1.: /9 fm, ,F— 4;, Which of the following is a FALSE statement? In a transverse wave the particle motion is perpendicular to the velocity vector of the wave. Not all waves are mechanical in nature. The speed of a wave and the speed of the vibrating particles that constitute the wave are different entities. Waves transport energy and matter from one region to another. A wave in which particles move back and forth in the same direction as the wave is moving is called a longitudinal wave. MSG; ‘\ Physics 125 Midterm Examination Winter 2010 Page 8 o i U ( 13. A supersonic aircrafi is flying parallel to the ground. When ‘34—— H) I fix the aircraft is directly overhead, an observer sees a rocket 9,“;‘1’ ‘ > i " I, «I " —\ ired (in the form of an explosion) from the aircraft. Ten /\ h? “fl” '9” " r 1 7 conds later the observer hears the sonic boom from the L 7” I/aircrafi, followed by the sound of the explosion 2.8 5 later. What is the speed of the aircraft expressed as a multiple of / the speed of sound? (mach number) x i / (a) 1.2 (b) 1.4 (d) 1.8 I ,/ (e) 2.0 ' (i ' N Mold A A‘ r. _)/ 41 6" t L" Maui as U, I c/ 8014ng _> 71qu f /2-3’L’ J zc) a/s / " . b” - 1 x ._ ~ 2,4,2 ; LIL—Y7 225%! if ; M49» r 2:; (2M (5" w L > to ‘ /¢’ 51%: £30519 a? 61’- _o , /0 ,' 9: 35‘ 6 (t 5" 7:57 I ‘4Aer: 0‘517 ‘ 1::«. Mac“ a: ...
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This note was uploaded on 04/01/2012 for the course PHYS 125 taught by Professor Mohamed during the Spring '08 term at Waterloo.

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Solutions_p125MidW10 - UNIVERSITY OF WATERLOO DEPARTMENT OF...

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