Physics 1B: Review for Final Exam Solutions

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Everett Murphy
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1 Physics B: eview for Finl Exm s Andrew Forrester June 6, 2008 In this worksheet we review mteril from the following chpters of Young nd Freedmn plus some dditionl concepts): Chpter 3: Periodic Motion Chpter 5: Mechnicl Wves Chpter 6: Sound nd Hering Chpter 25: Current, esistnce, nd Electromotive Force Chpter 26: Direct-Current Circuits. Dmped hrmonic oscilltion of chrged o. A spring of relxed-length L nd spring constnt k is ttched to the center of tletop of width W. An rm of negligile weight is ttched to the spring with one end immersed in viscous fluid tht provides dmping constnt for oscilltion of the spring. A smll o of mss m with chrge q insulted so the chrge does not escpe) is ttched to the spring nd held y you) so tht there is no force etween the o nd the spring. You let go of the o t time t = 0. ) Wht is the height y of the o ove the tletop s function of time? ) Using wht you know out electric fields, wht is the mgnitude of the electric field t the edge of the tletop point A in the 2D drwing) s function of time? c) Wht is the mgnitude of the electric field t the edge of the tletop fter very long time? ) y = yt)? The o strts off some distnce ove the equilirium position y eq of its oscilltion. Tht distnce is the initil mplitude of the oscilltion A 0. The nturl frequency of oscilltion is ω n = k/m, nd the decy rte for the mplitude of oscilltion is r = /2m, so the expression for yt) in these terms is yt) = y eq + A 0 e rt cosω t), where ω = ωn 2 r 2 = k/m 2 /4m 2. The initil phse φ 0 is zero since the oscilltion strts t its pek nd we re using cosine.) Now, wht re y eq nd A 0 in terms of the given quntities? At the equilirium position the spring is compressed y A 0 so tht the spring force pushing upwrds equls the grvittionl force pulling downwrds: ka 0 = mg. Thus A 0 = mg/k. Since the uncompressed length of the spring is L, the equilirium height is y eq = L A 0 = L mg/k. Therefore, yt) = L mg/k ) + mg/k ) ) e /2m)t cos k/m 2 /4m 2 t ) E = A? E = E = K q /r 2 = /4πε 0 ) q / [ W/2) 2 + y 2] [ Et) = q / {4πε 0 W 2 /4 + y t)] } 2 where yt) is given in prt ) c) A s t?

2 After long time, s t, the oscilltions die out nd yt) y eq = L mg/k, so E ) = q / {4πε 0 [W 2 /4 + L mg/k) 2]} 2. Mechnicl wve of vrile speed. Y&F 5.82) A deep-se diver is suspended eneth the surfce of Loch Ness y cle of length L tht is ttched to ot on the surfce. The diver nd his suit hve totl mss M nd volume V. The cle hs dimeter d nd liner mss density µ. The diver thinks he sees something moving in the murky depths nd jerks the end of the cle ck nd forth to send tnsverse wves up the cle s signl to his compnions in the ot. ) Wht is the tension in the cle t its lower end, where it is ttched to the diver? Do not forget to include the uoynt force tht the wter density ρ w ) exerts on him. ) Wht is the tension in the cle distnce x over the diver. The uoynt force on the cle must e included in your nlysis. c) The speed of trnsverse wves on the cle is given y v = F/µ. The speed therefore vries long the cle, since the tension is not constnt. This expression neglects the dmping force tht the wter exerts on the moving cle.) Integrte to find the time required for the first signl to rech the surfce. ) F diver? An unknown force suggests the use of free-ody digrm with Newton s second lw of motion. We my select our system to e the diver nd his suit), nd the force from the rope on the diver will e the tension F T tht we re looking for. Since there is no ccelertion, the net force equls zero. Thus F x = F T + F uoy F g = 0 nd F T = F g F uoy = Mg ρ w V g = M ρ w V )g = F T. ) F distnce x ove diver? We cn use the sme tctic s in prt ) ut declre the system to include the cle up to position x ove the diver. Agin, since F x = 0, we hve F T = F g F uoy = M tot g ρ w V tot g = M + µx)g ρ w V + πd/2) 2 x)g = M ρ w V )g + µ + ρ w πd 2 /4)gx = F T. c) Time t for signl to rech surfce? To simplify the mthemtics let s define nd such tht F T = + x. We hve v = F T /µ = µ /2 + x. We wnt to know t = t 0 dt. We cn relte v to dt y v = dx/dt, so t = t 0 L dx dt = 0 vx) = L dx µ nd letting u = + x, so tht dx = 0 + x du, t = µ +L du µ [ ] +L = 2 u = 2 µ [ ] u + L t = 2 µ µ + ρ w πd 2 /4)g [ M ρw V )g M ρ w V )g + µ + ρ w πd 2 /4)gL] 2

3 3. Sound wve interference. Y&F 6.84) Two loudspekers, A nd B, rdite sound uniformly in ll directions in ir t 20 C. The coustic power output from A is W, nd from B it is W. Both loudspekers re virting in phse t frequency of 72 Hz. ) Determine the difference in phse of the two signls t point C long the line joining A nd B, 3.00 m from B nd 4.00 m from A. ) Determine the intensity t point C from speker A if speker B is turmed off nd the intensity t point C from speker B if speker A is turned off. c) With oth spekers on, wht is the intensity t C? ) C? Since v = f λ, λ = v/f = 344 m/s)/72 Hz) = 2.00 m) = λ Speker A is 2 wvelengths 4π rd) wy from C nd speker B is.5 wvelengths 3π rd) wy. Thus the phse difference is φ = π,.k.. 80, so the signls re out of phse. Using less words nd more mth, we could clculte φ in the following mnner: φ = φ A φ B = kd A kd B = kd A d B ) = 2π/λ)d A d B ) = 2πf/v)d A d B ) = 2π72 Hz)/344 m/s) ) 4.00 m 3.00 m) = π ) I A nd I C? I A = P A /4πd 2 A = W)/4π4.00 m) 2 = W/m 2 ) = I A I B = P B /4πd 2 B = W)/4π3.00 m) 2 = W/m 2 ) = I B speker B off) speker A off) c) I C? The importnt conceptul issue is to note tht intensities do not dd; mplitudes of displcement or reltive pressure) dd, nd intensity is the squre of the mplitude. Also, since the signls re 80 out of phse, while one signl is t its mximum, the other is t its minimum negtive displcement or reltive pressure), so we must sutrct the mplitudes: I AB = I A I B ) 2 = W/m W/m 2 ) 2 = W/m 2 ) = I AB 4. Current nd resistnce etween concentric spheres. Y&F 25.64) The region etween two concentric conducting spheres with rdii nd is filled with conducting mteril with resistivity ρ. ) Show tht the resistnce etween the spheres is given y = ρ/4π)/ /). ) Derive n expression for the current density s function of rdius, in terms of the potentil difference V etween the spheres. c) Show tht the result in prt ) reduces to = ρl/a when the seprtion L = etween the spheres is smll. ) Show = ρ/4π)/ /). For simple geometries we use = ρl/a, where L is the length of the resistor long which the chrges trvel nd A is the cross-sectionl re perpendiculr to the direction of chrge trvel, ut this cse is not so simple. If we ssume tht chrge flows rdilly either outwrd or inwrd) when 3

4 this resistor is in use, then this sitution is s if we hd put mny resistors in rdil pttern, connecting the conductor of rdius to the conductor of rdius, with the resistors ll in prllel. Equivlently, we cn think of this sitution s if we hd plced mny thin, resistive sphericl shells concentriclly upon ech other, so tht they ct s resistors in series. Let s use oth of these pictures to mke two equivlent clcultions. Series: Although these concentric shells re perhps conceptully unfmilir, the clcultion is simple. For thin sphericl shell of infinitesiml thickness dr, dr is the length long which the chrges trvel nd A = 4πr 2 is the re through which they trvel, so the infinitesiml resistnce of the shell is d = ρ dr/4πr 2. Thus the totl resistnce is since resistnces in series dd or integrte) = d = ρ dr 4πr 2 = ρ dr 4π r 2 = ρ [ ] = ρ 4π r 4π ) Prllel: Although the picture of rdilly rrnged resistors is mde of conceptully fmilir resistors unlike the sphericl shell resistors), the clcultion is quite unusul. I ve ctully never seen nything written like this efore... I m just showing this to you for completeness. You should never hve to do nything like this... this is crzy.) d = ) = d ) = d ) d = = ρ ll ngles d ) = ρ dr d d = ) J = Jr, V )? V = I nd J = I/Ar) = I/4πr 2 ρ dr r 2 sin θ dθ dφ = ρ dr rdθ r sin θdφ ρ dr sin θ dθ dφ r 2 = ρ sin θ dθ dφ ) ) d = ρ ) sin θ dθ dφ ρ ) sin θ dθ dφ = ρ ) π 2π sin θ dθ dφ 0 0 ) [ cos θ] π 0 [φ]2π 0 = ρ ) + )2π 0) = ρ ) 4π = ρ 4π ) Jr) = V /) 4πr 2 = V 4πr 2 4π ρ ) = V ρ ) r 2 = V ρ )r 2 = Jr) c) Show ρl/a for L = smll enough. = ρ 4π ) ρ ) = 4π = ρl 4π = ρl 4π + L) ρl 4π 2 = ρl A where the pproximtion is true if L such tht + L. 4

5 5. Voltges in n C circuit with switch. Y&F 26.74) ) Wht is the potentil of point with respect to point in the figure elow when switch S is open? ) Which point, or is t higher potentil? c) Wht is the finl potentil of point with respect to ground when switch S is closed? d) How much does the chrge on ech cpcitor chnge when S is closed? ) V with switch open? Ech cpcitor ecomes fully chrged nd there is no longer current in the resistors. V = V V = E 0 = E = V ) Point higher potentil? Point is connected to the source nd point is connected to ground V = 0). Given E > 0, point is t higher potentil. c) V long fter switch closed? Chrge will lwys run through the resistors nd the cpcitors will ecome chrged ccording to the potentils cross the resistors. The current in the resistors is I = E/ + 2 ), so V = I 2 = E 2 / + 2 ) = V. d) Q of ech cpcitor fter switch closed? C : Q = C I ) C E = C E + 2 C E = C E ) = Q + 2 E C 2 : Q 2 = C 2 I 2 ) C 2 E = C 2 2 C 2 E = C 2 E ) 2 = Q Equivlent resistnce of cue frme of resistors. Y&F 26.92) Suppose resistor lies long ech edge of cue 2 resistors in ll) with connections t the corners. Find the equivlent resistnce etween two digonlly opposite corners of the cue points nd in the figure). We cn use the symmetry of the cue nd imgine chrge flowing to get rtio etween voltge nd current. Chrge flowing through should split evenly mong the three rnches it meets since there s nothing to distinguish them. Those three rnches, nd the three rnches meeting point, should ech hve current I/3. The sme rgument shows tht the six remining rnches hve current I/6. If we pick ny pth from to, we get Thus we hve V = I 3 + I 6 + I 3 = 5 6 I eff = V I = 5 6 = eff 5

6 We cn lso see tht, effectively, the ends of the three rnches ttched to re t the sme potentil since the resistnces, nd voltges cross them, re the sme. The lst drwing shows tht this implies tht the rrngement is effectively like three resistors in prllel, in series with six resistors in prllel, in series with three resistors in prllel. eff = = = = 5 6 6

Physics 1402: Lecture 7 Today s Agenda

Physics 1402: Lecture 7 Today s Agenda 1 Physics 1402: Lecture 7 Tody s gend nnouncements: Lectures posted on: www.phys.uconn.edu/~rcote/ HW ssignments, solutions etc. Homework #2: On Msterphysics tody: due Fridy Go to msteringphysics.com Ls: