123 - Exam 2 - Fall 2010 - solutions

123 - Exam 2 - Fall 2010 - solutions - g0 (U +10fl3 (15...

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Unformatted text preview: g0 (U +10fl3 (15 pts) Problem 1: Multiple choice conceptual questions. Choose the best answer. Fill in your answers on the bubble sheet. 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. 1.8. You have two balloons, one filled with air and one filled with helium. If you put both balloons into a tub of liquid nitrogen, which one will end up with the smallest volume? ,7 , A9 . Q; the air balloon \ogccuuik m0“; -\" A 0,“ (,JJ \ e W‘ \ASK V. .jfl b. the helium balloon (Ni 0).?) NM? v} gums ) c. they will end up with the same volume \ \\CL\.\\ a). The second law of thermodynamics is a statement of: a. conservation of energy b. conservation of (regular) momentum 0. conservation of angular momentum d. conservation of mass/volume ’- \ Lust-l +0 Hm") 521%». Ch.) 4; ‘40‘\' ) 74'; 3Q" probability 3:4 (5 m4 \W\ assflo x . \A/ \x‘yL\\l. \‘( Frm (his) As I'm taking data in my lab, I often take data fro - i ht detector and average the data for 1 second for each data point. You can assume that this represents data figndOfiO photons, for example. Suppose I decide to average the data for 2 seconds per point instead (collectinga2fl‘9m0 hotons, for example). How much better is my signal—to—noise ratio likely to be? Hint: Think about the statistic?5 uctuations (the “n01se ")‘likely to occur in each case. a. the same . .Clefid‘gfi : \eco VA 3 @ «l2 times better £HCNJW6¢ WOO I . 103°.._°°°. . \eaofi c. 2times better _ \moooo N " Na)“— . g/N I a... 4000 d. 4 t1mes better 10% \ fi- . 000 e. 8t1mes better “Jam 0* g,” 3 13;; .I {2“ In the “ladies belt demo” (the belt was like a “closed-closed” string), suppose the fundamental frequency is seen at 500 Hz. What frequency will have five antinodes? :. fggOHz yam.“ hams-mg, <2; 5238 c. 1500 WC; 1 53 . 3000 Hz /).2 rMZuwx VA] éfép‘t‘fi‘“ Whe - “wave” is actually a finite duration pulse, the group velocity IS the speed at which: the overall pulse envelope propagates ~ ( b. all individual frequency components propagate wittu’e ‘3 {‘0 4m)" we ‘ V ‘1 \ 4 the average frequency component propagates —»> 4133 comm "* “W V‘“°"“ “1 q" :3 none of the above more than one of the above \ h 0‘ W‘ngva t.) {qu05% C. d. C. You should have found the wave speed of waves on your slinky to be directly proportional to the slinky’s length. (This was true for both longitudinal and transverse waves.) Compare the time it takes waves to do a round-trip path when the slinky is stretched to 5 feet compared to when it is stretched to 10 feet. ‘ tm >t10fi at : iv 5« If x 0H9 \I We ‘4’ M 0W5” J‘ 31W“ H“ SW" tSft m tloa I _ C. t5fi <t10ft (\‘Q “QM cyamqg'iln I 5 Complete the following based on our in-class discussions: To localize a wave in space, you need lots of a. amplitude b. energy 0. patience , _ x d. phase u. 0A9 A“ k‘L‘m’Q "i "WWW ' “1M V‘ch C9 wavenumbers \ 6/} g )1 fYQ/qoaewtg - The sound of. a trumpet playing a note is qualitatively different than the sound of a flute playing the same note. Why is that? a. The two notes have different amplitudes. b. The two notes have different durations. c. The two notes have different fundamental frequencies. (1. The two notes have different phases. The two notes have different strengths of harmonics. Olfieu‘i UMP xiWNQ “’5 d‘W‘b“ :Fow-‘V C°‘H‘°“d§ Thermo Exam iL— pg 2 [‘1 CosOswa’C +51) 1.9. The wavefunction for a particular wave on a string is given by: f(x,t) = 4 cos(2x —— 6t + 5). You may assume that the numbers in that equation and in the answer choices below are all given in terms of the appropriate SI units. What is the amplitude? a. 2 e. 6 i. 211/5 b. 3 f. 211/2 j. 211/6 6,) 4 A g. 211/3 (1. 5 h. 211/4 1.10. Same equation. What is the wavelength? a. 2 2.”— e. 6 i. 211/5 b. )6 TL (9 211/2 j. 211/6 0. 4 g. 211/3 d 5 L > 9/ h. 211/4 1.11. Same equation. What is the wavenumber (k)? a. 2 e. 6 i. 211/5 11/ 3 f. 211/2 j. 211/6 0- 4 g. 211/3 d- 5 h. 271/4 1.12. Same equation. What is the period? a. 2 7{"7/ e 6 i. 211/5 b. 3 «r: N f 211/2 211/6 0. 4 i (0 g 211/3 1 d. 5 U3 h 211/4 1.13. Same equation. What is the angular frequency (a) 9 a 2 Q 6 i. 211/5 b. 3 f. 211/2 j. 211/6 0. 4 g. 211/3 d- 5 h 211/4 1.14. Same equation. What is the phase ((15)? a. 2 e. 6 i. 211/5 b. 3 f. 211/2 j. 211/6 0. 4 g. 211/3 @ 5 h. 211/4 1.15. Same equation. What is the wave’s speed? a. 2 e. 6 i. 211/5 (:1) 3 f. 271/2 j. 211/6 4 g. 211/3 d- 5 (g g h. 211/4 -' If f V ‘ \/ "‘ ’12., 2" J \‘ .4 .u u 051$“ 1915” L! (a..- VNC) I (N 3“ Thermo Exami— pg 3 (1 1 pts) Problem 2. (a) A car (traveling at 44 m/s) is chasing you on your bike (traveling at 12 m/s). The car driver honks her horn and emits a tone of 500 Hz. Use 343 m/s for the speed of sound. What frequency do you hear? x! i V0 (I: (F V t ‘f ' VviKL 4w blccw‘a’“ ‘i’bSVW‘J :0" 3‘ e . I in“ [a c“ l mvws 4'7” 4 r a in {NM 2"” ‘1‘“ Vk MV‘ 1 “Ski/5 # I} ‘3’ WW 6 L B chow) “OOH f Ll“! Helium“? 3 " S . ’3 5 a (5% I; '" K 7 one \a4fiJ I’ll Cau‘k‘ Sanc'Qv ".3 F «mum; $ 3““ “m' ' 50M (tweak; mgwlr HNWJQ QMMNJV fig? 0% W: «>va bcjtfllfiwh‘ 3i1t¥~+g V9 X (b) The animated gif I showed in class where the group and phase velocities were in opposite directions had the following dispersion relation: 0) = —1/k2. (You can assume that the number “1” in the equation has the appropriate units to make 0) be in rad/s and k in rad/m.) It was made up of 21 frequency components centered around k = 1 rad/m. Find the group and phase velocities of the pulse. Ln}: 2. W ‘(01 A LIMP w V i _ (wins?) -. L 3” '\\r ’ "k <;, “‘ til:- ‘4’, ’ng VgToup: pal/“2” ( e2 2 f; i may )G'L‘ QLSD LW‘UCAL‘52 or”! Vphase Thermo Exam ’11: pg 4 5 «+9 (10 pts) Problem 3. Guitar players often tune their instrument using harmonics. Imagine that I use an electronic tuner to tune my low E string to 164.814 Hz, precisely the correct frequency for an equal temperament scale referenced to the standard frequency of 440.000 Hz for the A above middle C. Now I tune my next string, the A string—which is 5 half-steps higher in frequency, a musical "fourth"— using harmonics. I do this by lightly touching the E string 1/4 of the way from the end of the string and lightly touching the A string 1/3 of the way from the end of the string and adjusting its frequency so that the beats go away. (a) What is the frequency ratio of the two notes and why does this tuning method work? if ‘ift ‘1 L! (mamas) ’l‘OVC‘A‘; \i \IJN'l 410‘ Low mm. 14> 4’44 kumgzv 7. A ‘hudvx DA V3 WAVI 8 N hwy-AW“ L/ o)‘k \ka —‘:u( (is V‘ : QJASW; a a , «a (war-I m) c ’3 (19%;qu J19 gel-{2.53 f; TUhf-qfi awq7 w.- w um»— " q ‘ 1,“; Wt?“ a I ~11 i “m~v—.w.-m..mgm.w(wmfiw.u (b) What is the difference in frequency between an A tuned that way and an A tuned according to the equal temperament scale? EqUhi “"lr‘EMVIEJVW‘i‘ he.» 5 lixG\{—'$l"§¢j \’\:3l\g/ {L} WM : Fifi); x lwaiil‘l Hi > ZloHt 9’3 (,e ‘3) (A 5““M W 05am gfivi iWfiv> “i. "i 16’“? of @‘Hal‘l2\ Thermo Exam ’L— pg 5 W 3r r; m?” (11 pts) Problem 4. Two small speakers emit MW 3, both with outputs of 2 mW of sound power. (a) Assuming the two waves are incoherent so that the intensities add, determine the intensity (W/mz) and sound level (dB) experienced by an observer located at point C. 4.00 m \ \ ..........___..._.._..~_.h.___-, .,_....’;.(, \l 36/ intensity = W/m2 I (L7 2 l o \v t in, > "3' * - \o t (156* W” " ‘“ "fr-J.- \"L. <3 r Wm ; 7 l . S (0 (ix i) 5W sound level = dB (b) Assuming the two waves are coherent, so that the amplitudes add (to produce a pattern of minima and maxima), what is the lowest frequency that will produce a maximum at point C? The two speakers are governed by sinusoidal voltage sources that have the same frequency and are in phase with each other. Use vsound = 343 m/s. . w = V > Map/we. " (Haul/bx in-NZMI ““ n A “A QC \th (M1 «~3 n \ f A ea 3 n ‘5 b m A 1% m " I 4 Set Thermo Exam ’L— pg 6 (12 pts) Problem 5. Add these two cosine functions together using complex number techniques to determine the amplitude and phase of the resulting function: f1 = 3cos(4t + 5), f2 = 4cos(4t + 6), and thereby write down the resulting function. The numbers "4" and "6" have units of rad/s and radians, respectively. Hint: be careful as to whether your calculator is in degrees or in radians mode. (If you know the method, you don’t necessarily have to write down any complex numbers to solve this problem.) J diggusmfi S“ CU.“ A§§wiU QM {Aqétu 4 “Sum u-i S;V\-\J\Q:¢>€4 “Fume/4‘5“; 7‘: [mi-- / J i J \ C“ C (kg-«g Ra pa in; Can/b $315 A vchs- \(j‘ 2 B L§ PM} : g ( QKQ’gb p _ a WW “"‘Nwagif. Jj _ L} LQJ‘g 2 \el 3‘4 3.8’ > z w 4 pI {I +va w-‘u ‘09‘ Swazi} “Ks ’IL‘R Ai‘i A1171 I“ MM? 2 30: 23am- manner .1 9,671 Yams 1 ’5 gig was"ch magmas" ; “3,97%! *\}zw is g); [I ; 19M I’M—w” weika : {kt/em" + 3am” 1t;th V .4 72.4%?” gm Wtr (first) ...,.... ._.vw-""" .; t."w“gummy”-..u.‘gm‘,M_uN" WWW“: w. (W: fl 77053) LIrifl’L rcwrt‘aj Mae: 4m: M92 m v—"w—MN--—~-—-~m_.- , - e...i...v..7,.t—.‘ “mm.” m —- »- Amplitude = Phase = radians, = degrees Resulting function f(t) = f1(t) + f2(t) = Thermo Exam ’1’; pg 7 \/ \f l L (8 pts) Problem 6. A sound wave travels from air (v = 343 m/s) into a small lake (v = 1493 m/s), coming from a source far from the lake, and directly above it, so that the wave crests are essentially parallel to the water as it enters. This is thus a 1D situation, where the same equations we derived for waves on a string are applicable. (a) What fraction of the incident wave’s amplitude is reflected? i «xvi 1 N33122:? is :T “a Y‘" x” zuzflkkcté “K ° VIM/7. (b) What fraction of the incident wave’s amplitude is transmitted? Hint: don't be surprised if this result. . .well, surprises you. a M 1 10am) ,1”- ty \ZTV; 77w)“qu ’ , \ 5t. \) ‘l‘rw‘j “NH/0‘) “NOW I {V(( u’ ['\$I'\) "DAD mofk Ovmwlllu "I (D fl I S \0 3‘91, me {A {A 414,, 'P\-;S‘l flack ( (Sr) 71“ 5 1‘, I er 75 gm CSAQNLEB , L \ 5W ()\QM+,{‘||6)+Q This keg;ng \giLcoNkk “MW “RM (0) What fraction of the incident wave’s power is reflected? “land N1 yoqu rl T; IQLKleeél . 3127. 4 «Amps retrth (d) What fraction of the incident wave’s power is transmitted? T= \‘WL "’@ 6Q,®7/2 “it Q/mj'y l! elrwwxé’lu‘)? «hm,& Jews Lem n ovs we Thermo Exam ’L— pg 8 (12 pts) Problem 7. This equation: has a solution (meaning a function for which the equation is true) of the form x(t) = Ae‘m cos(a)t) . That solution represents, for example, a spring with an oscillating mass hanging from it, where the amplitude of oscillation continually decreases due to air resistance. 1 (—l+ia)) By representing x(t) as a complex exponential, x(t) = Ae T and plugging it into the equation, find what rand a) must be in terms of k, m, and 7, to make the equation true. (Yes, this is the same as your homework problem HW 16-5. Demonstrate that you can do the problem, don't just quote he answer to that problem.) * aran )(QW/Xc ’ 8 40%“) e: 1 (“wire , > 0‘ (I 4(,J7—J'\(W - 01* »<,;\I~J&tlr-’) Q“ R, ,r-)-uv—>) EL ’ m» " vi If M) “\v: ‘ ' LW)\ “'9: (ail/W”) /V\ .. .../_\ \lAJ C, k" W‘ ‘Qlus m‘k) %Mm+\m L 71f air 1 ) ,_ ic’ I —» —' Jaw a WW) - wt 7/ c : L 5/ You a, ’L ,.. , W K- , k m ’L " L IN — r m’i" m ‘ 4,1 ’l’ Thermo Exam‘l — pg 9 (15 pts) Problem 8. In my lab I have a voltage “function generator” which can I. .1 9‘5 l E; i. .x. produce triangular—shaped voltage patterns such as the one shown. The x-axis is .. i .a 'r l I : t seconds; the y-axis is volts. (You can ignore units for the rest of this problem.) From ‘1‘ ,’ l 2‘ joy“ , ‘1‘, t it 1‘ . '.i 0 to 0.5 s it sends out a signal that isf(x) =x. From 0.5 s to 1.0 s it sends out a signal I! 0i3 ,i f ‘3 that is /(x) = 1 —x. Then the signal repeats. 5' ‘1 '1 l I! i ' .' 3-, i .5 i‘ i l i g a i g a '1; .-' \. 1 a 5 i z oh i i v' ‘ 3 " ,z’ '1 Determine the Fourier coefficients of this function and write f(x) as a sum of sines, ,I‘ ii i '1 (I \x cosines, and a constant term, as applicable. If any of the Fourier coefficients have i f 0.1" j obvious values, state your reasons why it/they are obvious. 1" i- if ix l ‘ i 1 ‘i' 1 . —4 —2 ' 2 4 Potentially useful integrals: . cos(27mx) sin(27rnx) s1n 27mx dx = ——————— J‘cos 27mx dx = —— :_ I ( ) 27m ( ) 27m L / . 2 ' 2 ‘ 2 ' 2 Ix s1n(27mx)dx = ——x COS( mm) +—Sm( 2mg) Ix cos(27mx) dx = —cos( 2”?) + —xsm( mm) 27m 47: n 47: 11 27M Note: although you should recognize that sin(rm) and sin(2n1t) = 0 for all integer values of 11, you, can leave your answers in terms of cos(nn) and cos(2mt); that is, you don't have to work out the pattern of your answers in terms of odds/evens or (-1) to some power. (Q. die “(55:3 (0. an \9j tamj) Ewe-“~ch toms 0“?" WM L qurw ’3; g {(flcsi t AV b I 0 _ [L k/MNG’"" M" Casimx A x CHEW” 4 II.L l/ 2 >4 1 L l \ 1 15" «a 17ml 2 Cog—«es, i X5 at» 3.x} 1‘1th \ __ C05 21”“7‘ 5‘} 2qn‘1 rulf‘jflV WM, (/“TL/ \ 0 2n“ ‘1'“ h if), 0 VT’N MN I]L IL WW ” O :0 3‘ O h 2 (gsfiflrl Q33 11in ~<Qyfin Vi’fllk'L Lien?” V\ L 2mm“ x gas'bm “i Embglngwg anofih L / ck {his 7 [7— 1‘ n a W‘ “ 5‘ ‘- Eff/b gxmphve.) mQrA' (3932134 = i KKVJCk<5 , 3) ‘9’“ WWW Thenno Exam :1): pg 10 r) g H Cg“ 72K 0 (6 pts, no partial credit) Problem 9. Heat is added to 4 moles of m» natomlchdeal gas at 300K, while keeping its pressure constant as shown in the diagram. Heat comes in from a reservoir kept at 0—Krw lch cases the temperature of the gas to increase to 700K as shown in the P—V diagram. What was the change in entropy of the universe during this process? (Hint: find the change in entropy for the gas and for the reservoir separately, then add them together. By the 2"d Law, your final answer must be positive. You can assume that the heat lost by the reservoir is equation to the heat added to the gas.) 51.6(‘J‘3‘i P T . AS I " {El—,ng a “CV, 9m “54., A B 034$ / l T 869% : :3‘ 2 Q" \ res. Ag rtwoacr Z Tm; , 1 r l\3 ‘\;\3 W R (/ __\ C Thermo Exam 1,— pg 1 1 (4 pts, no partial credit) Extra Credit. Suppose you attach a rope (mass m, length L) to the ceiling and let it dangle down so that it just barely reaches the floor without touching. How long will it take for a transverse pulse to travel fiom the bottom of the rope to the top? Put your answer in terms of m, g, and L. Hint: You can do this in two steps. First, pick a point on the rope a distance x up from the bottom of the rope, and draw a free-body diagram for that point. There's some weight (but not all the weight) pulling down, and some tension pulling up. That should give you tension as a function of distance. You may assume that the rope’s linear mass density will not vary with height. You already know how the wave speed depends on tension and linear mass density, so you should then be able to figure out the wave speed as a function of distalnce. Then, use that information and some calculus to figure out the answer to the problem. «91% I M l l at. (-~ New CK. Omi .44,va j 4} fl 9"?“ \(L. ;3 “is; 'pscrimél {EN 6\‘ mom 6: ;u g x K (07. i110 o A} A“: 6» f0 f? L (rc"& I‘Vlv \‘ , 5k:- ' 7‘“ " “Q M m a l-exc‘iar a 'T‘TU “VA ° 3 I J \ Thermo Exam ’1.— pg 12 ...
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123 - Exam 2 - Fall 2010 - solutions - g0 (U +10fl3 (15...

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