PHY138 Waves Test Fall Solutions

Transcription

1 PHY38 Waves Test Fall 5 - Solutios Multiple Choice, Versio Questio simple harmoic oscillator begis at the equilibrium positio with o-zero speed. t a time whe the magitude of the displacemet is ¼ of its maximum value, through what phase agle has the oscillator moved? () (.3 Nπ) radias (N a iteger) * (B) (Nπ ±.53) radias (N a iteger) (C) 75.5 degrees (D)(4.7 N 8) degrees (N a iteger) (E) (Nπ +.3) radias (N a iteger) Reasoig: If the oscillator begis at equilibrium, x=, the we ca write x=si(ωt). We are ot sure if it is iitiall movig up or dow, so we do t kow if is positive or egative. Set x=/4, solve for ωt, which will be the phase agle the oscillator has moved through (sice ωt= at the begiig). /4=± si(ωt), ωt=si (±.5). The aswer o our calculator is.53 radias, but a look at the si(θ) curve shows that Nπ ±.53 for N=, ±, ±, etc also is a solutio to si (±.5). Questio What is the magitude of the phase differece, i radias, betwee the displacemet wave ad the correspodig pressure wave i a siusoidal soud wave? () (B) π/6 (C) π/4 *(D) π/ (E) π Reasoig: s discussed i class ad the text, pressure is at a maximum whe logitudial displacemet is zero. This represets a phase differece of π/. Questio 3 Radar ivolves usig a electromagetic wave to detect a object b observig the eerg reflected from that object. ssume the target object is a perfectl reflectig circular disk of area,. The receivig atea has a collectig area of. Both the emittig ad receivig ateae are at a distace, d, from the target object. What is the ratio of the received power to the power origiall radiated b the emittig atea? () (B) (C) *(D) (E) 4 4 6π d 6πd 4πd 8π d πd Reasoig: Set the power emitted b the emittig atea to be P. It spreads out i all directios, so the itesit a distace d awa is I =P/(4πd ). This is the itesit at the circular disk. The power that is reflected b the circular disk is P =I (), where is the area of the disk. ssumig it spreads out i all directios, the itesit a distace d awa is I =P /(4πd )=P/(6π d 4 ). This is the itesit back at the receivig atea. The power that is collected b the receivig atea is P =I (), where is the collectig area of the receivig atea. So P =P() /(6π d 4 )= P /(8π d 4 ), ad the ratio P /P= /(8π d 4 ), Questio 4 trumpet plaer stads o a rollig platform, which moves at a costat velocit towards a large statioar brick wall. He blows ito the trumpet, producig a soud with costat frequec of f = 56 Hz, as measured whe the trumpet ad a observer are both at rest. The trumpet plaer hears the emitted soud as well as the echo of the soud which is reflected from the wall. He hears a beat frequec of f beat = 3. Hz. What is the speed of the trumpet plaer? ().5 m/s (B).5 m/s *(C). m/s (D) 4. m/s (E) 33 m/s Reasoig: Call the speed the trumpet plaer moves v t, the beat frequec he hears is f b, ad the speed of soud v=343m/s. We eed to solve for v t. The wall is statioar ad the trumpet is movig, so the frequec the wall receives is foud from the Doppler equatio for a approachig source: f =f /( v t /v). The wall reflects this soud, so it acts like a source which is statioar ad emittig with frequec f. The trumpet plaer is movig toward this source, so the frequec he hears is foud from the Doppler equatio for a approachig observer: f =f (+v t /v)=f (+v t /v) /( v t /v). This is slightl higher tha f, so the beat frequec heard b the trumpet plaer is f b =f f. Or, f b = f (+v t /v) /( v t /v) f. Everthig i this equatio is kow except for v t. Solvig for v t takes a bit of algebra, ad the aswer is v t =vf b /(f+fb)=(343)(3)/( 56+3)=. m/s. Page of 6

2 PHY 38Y Waves Test Detailed Solutios 5 Questio 5 stadig wave is oscillatig at 95 Hz o a strig, as show i the figure. What is the wave speed? () 9 m/s (B) 9 m/s (C) 343 m/s *(D) 38 m/s (E) 57 m/s Reasoig: The wavelegth of a stadig wave is twice the distace betwee adjacet odes. I this case, it ca be foud from the diagram to be λ(/3) 6 cm =.4 m. The speed is foud from v= λf=(.4)(95)=38 m/s. Questio 6 form of soud-proofig is a fie wire mesh which is held at a fixed distace from a flat wall. Whe souds waves are ormall icidet o the wall, the first ecouter the mesh. bout half of the soud itesit is reflected, ad half is trasmitted. The trasmitted soud waves ca the travel the distace, d, reflect off the wall, travel the distace d agai, ad the combie with the origial reflected soud from the wire mesh. If the two soud waves are exactl out of phase at this poit, the will destructivel iterfere, reducig the total reflected soud itesit. If d =.54 cm (oe ich ), what is the miimum frequec for which the soud-proofig will work properl? () 35 Hz (B) 4 Hz *(C) 338 Hz (D) 675 Hz (E) 3,5 Hz Reasoig: The reflected wave from the wall travels a total extra distace of d further tha the reflected wave from the mesh. (This is similar to what happes i thi-film iterferece.) The phase dela will be kx=π for the first destructive iterferece, where x=d is the path differece. Solvig for k we have k= π/d. The wavelegth is λ= π/k=4d. Recall that the speed of soud waves is v= λf. Solvig for frequec, f=v/λ=v/4d=(343 m/s)/(4.54 m) = 3375 Hz. (This questio is similar to the suggested Kight Problem.65.) Questio 7 fish tak whose bottom is a mirror is filled with water to a depth, d. small fish floats motioless, a distace uder the surface of the water. The idex of refractio of the water is. What is the apparet depth of the reflectio of the fish i the bottom of the tak whe viewed at ormal icidece? d d d d () (B) (C) (D) *(E) Reasoig: I order to observe the reflectio of the fish, the light ras must first reflect from the mirror s surface, the pass through the water/air surface of the tak. First surface: The object is the fish. It is a distace s=d from the mirror, so the image is a distace s =s=d below the mirror. Secod surface: the object is the image from the first surface, which is (d ) below the bottom of the tak. So s=d+ (d )=d. To fid the apparet depth we assume the object is immersed i water with =, ad the observer is i air with =. The apparet depth is s =( / )s = (d )/. (This questio is similar to a assig masterigphsics problem, How Deep Is the Goldfish from the Chapter 3 Problem Set.) Questio 8 ra of light is icidet at agle θ i =3. o the surface of a liquid from below. The refractive idex of the liquid depeds o the wavelegth of the light, ad is give b =.5 3 λ, where λ is i aometers (m). ir with idex of refractio =. exists above the liquid. What is the miimum wavelegth for which total iteral reflectio ca occur? () m (B) 47.7 m (C) 55.4 m (D) m *(E) m Reasoig: Let s set C=.5 3, so that =C λ. For the miimum wavelegth, we will wat the ra to be exactl at the critical agle, so that θ c =3.. The total iteral reflectio equatio is si θ c = / =/C λ. Solvig for λ gives λ=/(csi3)=754.8 m. Page of 8

3 PHY 38Y Waves Test Detailed Solutios 5 Multiple Choice, Versio B Questio What is the magitude of the phase differece, i radias, betwee the displacemet wave ad the correspodig pressure wave i a siusoidal soud wave? (B) π/6 (B) (C) π (D) π/4 *(E) π/ Reasoig: s discussed i class ad the text, pressure is at a maximum whe logitudial displacemet is zero. This represets a phase differece of π/. Questio simple harmoic oscillator begis at the equilibrium positio with o-zero speed. t a time whe the magitude of the displacemet is ¼ of its maximum value, through what phase agle has the oscillator moved? *() (Nπ ±.53) radias (N a iteger) (B) (.3 Nπ) radias (N a iteger) (C) (4.7 N 8) degrees (N a iteger) (D) 75.5 degrees (E) (Nπ +.3) radias (N a iteger) Reasoig: If the oscillator begis at equilibrium, x=, the we ca write x=si(ωt). We are ot sure if it is iitiall movig up or dow, so we do t kow if is positive or egative. Set x=/4, solve for ωt, which will be the phase agle the oscillator has moved through (sice ωt= at the begiig). /4=± si(ωt), ωt=si (±.5). The aswer o our calculator is.53 radias, but a look at the si(θ) curve shows that Nπ ±.53 for N=, ±, ±, etc also is a solutio to si (±.5). Questio 3 trumpet plaer stads o a rollig platform, which moves at a costat velocit towards a large statioar brick wall. He blows ito the trumpet, producig a soud with costat frequec of f = 56 Hz, as measured whe the trumpet ad a observer are both at rest. The trumpet plaer hears the emitted soud as well as the echo of the soud which is reflected from the wall. He hears a beat frequec of f beat = 3. Hz. What is the speed of the trumpet plaer? ().5 m/s (B).5 m/s (C) 33 m/s *(D). m/s (E) 4. m/s Reasoig: Call the speed the trumpet plaer moves v t, the beat frequec he hears is f b, ad the speed of soud v=343m/s. We eed to solve for v t. The wall is statioar ad the trumpet is movig, so the frequec the wall receives is foud from the Doppler equatio for a approachig source: f =f /( v t /v). The wall reflects this soud, so it acts like a source which is statioar ad emittig with frequec f. The trumpet plaer is movig toward this source, so the frequec he hears is foud from the Doppler equatio for a approachig observer: f =f (+v t /v)=f (+v t /v) /( v t /v). This is slightl higher tha f, so the beat frequec heard b the trumpet plaer is f b =f f. Or, f b = f (+v t /v) /( v t /v) f. Everthig i this equatio is kow except for v t. Solvig for v t takes a bit of algebra, ad the aswer is v t =vf b /(f+fb)=(343)(3)/( 56+3)=. m/s. Questio 4 Radar ivolves usig a electromagetic wave to detect a object b observig the eerg reflected from that object. ssume the target object is a perfectl reflectig circular disk of area,. The receivig atea has a collectig area of. Both the emittig ad receivig ateae are at a distace, d, from the target object. What is the ratio of the received power to the power origiall radiated b the emittig atea? (B) (B) *(C) (D) (E) 4 4 6πd 6π d 8π d 4πd πd Reasoig: Set the power emitted b the emittig atea to be P. It spreads out i all directios, so the itesit a distace d awa is I =P/(4πd ). This is the itesit at the circular disk. The power that is reflected b the circular disk is P =I (), where is the area of the disk. ssumig it spreads out i all directios, the itesit a distace d awa is I =P /(4πd )=P/(6π d 4 ). This is the itesit back at the receivig atea. The power that is collected b the receivig atea is P =I (), where is the collectig area of the receivig atea. So P =P() /(6π d 4 )= P /(8π d 4 ), ad the ratio P /P= /(8π d 4 ), Page 3 of 8

4 PHY 38Y Waves Test Detailed Solutios 5 Questio 5 form of soud-proofig is a fie wire mesh which is held at a fixed distace from a flat wall. Whe souds waves are ormall icidet o the wall, the first ecouter the mesh. bout half of the soud itesit is reflected, ad half is trasmitted. The trasmitted soud waves ca the travel the distace, d, reflect off the wall, travel the distace d agai, ad the combie with the origial reflected soud from the wire mesh. If the two soud waves are exactl out of phase at this poit, the will destructivel iterfere, reducig the total reflected soud itesit. If d =.54 cm (oe ich ), what is the miimum frequec for which the soud-proofig will work properl? (B) 4 Hz (B) 35 Hz (C) 3,5 Hz *(D) 338 Hz (E) 675 Hz Reasoig: The reflected wave from the wall travels a total extra distace of d further tha the reflected wave from the mesh. (This is similar to what happes i thi-film iterferece.) The phase dela will be kx=π for the first destructive iterferece, where x=d is the path differece. Solvig for k we have k= π/d. The wavelegth is λ= π/k=4d. Recall that the speed of soud waves is v= λf. Solvig for frequec, f=v/λ=v/4d=(343 m/s)/(4.54 m) = 3375 Hz. (This questio is similar to the suggested Kight Problem.65.) Questio 6 stadig wave is oscillatig at 95 Hz o a strig, as show i the figure. What is the wave speed? (B) 9 m/s (B) 9 m/s *(C) 38 m/s (D) 343 m/s (E) 57 m/s Reasoig: The wavelegth of a stadig wave is twice the distace betwee adjacet odes. I this case, it ca be foud from the diagram to be λ(/3) 6 cm =.4 m. The speed is foud from v= λf=(.4)(95)=38 m/s. Questio 7 ra of light is icidet at agle θ i =3. o the surface of a liquid from below. The refractive idex of the liquid depeds o the wavelegth of the light, ad is give b =.5 3 λ, where λ is i aometers (m). ir with idex of refractio =. exists above the liquid. What is the miimum wavelegth for which total iteral reflectio ca occur? (B) 47.7 m (B) m *(C) m (D) 55.4 m (E) m Reasoig: Let s set C=.5 3, so that =C λ. For the miimum wavelegth, we will wat the ra to be exactl at the critical agle, so that θ c =3.. The total iteral reflectio equatio is si θ c = / =/C λ. Solvig for λ gives λ=/(csi3)=754.8 m. Questio 8 fish tak whose bottom is a mirror is filled with water to a depth, d. small fish floats motioless, a distace uder the surface of the water. The idex of refractio of the water is. What is the apparet depth of the reflectio of the fish i the bottom of the tak whe viewed at ormal icidece? d d d d (B) (B) (C) (D) *(E) Reasoig: I order to observe the reflectio of the fish, the light ras must first reflect from the mirror s surface, the pass through the water/air surface of the tak. First surface: The object is the fish. It is a distace s=d from the mirror, so the image is a distace s =s=d below the mirror. Secod surface: the object is the image from the first surface, which is (d ) below the bottom of the tak. So s=d+ (d )=d. To fid the apparet depth we assume the object is immersed i water with =, ad the observer is i air with =. The apparet depth is s =( / )s = (d )/. (This questio is similar to a assig masterigphsics problem, How Deep Is the Goldfish from the Chapter 3 Problem Set.) Page 4 of 8

5 PHY 38Y Waves Test Detailed Solutios 5 Log swer Questio wrog result used correctl i subsequet parts icurs o further pealt. correct method givig the correct aswer gets full marks. The questio is worth 36 poits. PRT (4 poits) While he is daglig, set the upward force of the bugee equal to the weight of the ma. The weight of the ma is Mg. The upward force of the bugee is foud from Hooke s Law: F= kx. I this case x is dowward, ad is equal i magitude to the ew legth, L mius the ustretched equilibrium legth, L. Defie positive to be upward, so x= (L L), so the force is F=k(L L). Settig the forces equal: Mg = kx Mg=k(L L) Mg=kL kl kl =kl+mg L = L + Mg/k Part B ( poits) The costat speed of the pulse is v pulse =L /t pulse, so t pulse =L /v pulse. Ts v pulse =. The legth of the bugee is L, so the time is foud from µ The mass desit of the bugee is its total mass divided b its curret legth: µ=m/l. The tesio i the bugee is Mg t L µ pulse = = L = v pulse Ts L m MgL ml t pulse = Mg Note: a reasoable simplificatio of the above form for the fial aswer gets the full two poits. Part C ( poits) There are several was to solve this problem. correct method givig the correct aswer gets full marks. Here are two methods: Method. Coservatio of Eerg Page 5 of 8

6 PHY 38Y Waves Test Detailed Solutios 5 t the top of the motio, his kietic eerg is zero, ad the bugee is ustretched, so his total eerg is just his gravitatioal potetial eerg. Let s set height= to be the poit whe the bugee cord is full exteded. So at the top of the motio, height=l max. E top = mgl max t the bottom of the motio, his kietic eerg is also zero, ad his height is also zero as we have defied it. ll of the eerg of the sstem is cotaied i the stretched bugee, which is a distace L max L awa from its equilibrium. So E bottom = ½ k (L max L). B coservatio of eerg: E top = E bottom mgl max = ½ k (L max L) MgL max = k L max kl max L + kl kl max (Mg + kl) L max + kl = Use the quadratic equatio to solve for L max : L max Mg + kl ± = (Mg + kl) k 4L k L max = Mg + L ± k M g k MgL + + L k L Mg Mg Lmax = L ± + L k k Numericall, L max = 7.5 ±.75 = 5 m or 4.5 m. Phsicall, L max must be loger tha 5 m, so the 5 m solutio is the ol oe that makes sese. This should have two sigificat figures, so studets ma either aswer 5 m or 5. m. Either wa to displa sigificat digits is acceptable. L max = 5 m Method. Simple Harmoic Motio with specified iitial coditios We kow from Part that L = L + Mg/k = 5 + (85)(9.8)/68 = 7.5 m. Whe the bugee cord is exteded, the motio will be Simple Harmoic Motio (S.H.M.) of a vertical oscillator, = cos( ω t + φ ) with the equilibrium, =, whe the bugee has a legth L. Let s defie to be positive upward. The agular frequec k 68 will be ω = = = rad/s. m 85 If we ca determie the positio,, ad velocit, v, at the begiig of the S.H.M. motio, we should be able to solve for the amplitude,, ad the determie the maximum legth of the bugee cord. (3 poits) Page 6 of 8

7 PHY 38Y Waves Test Detailed Solutios 5 t the begiig of S.H.M., =7.5 5 =.5 m, the bugee has just begu to stretch. bove that, the ma has bee free-fallig a distace of 5 m, startig from rest up o the bridge. From kiematics of costat acceleratio, we kow that v =v ± ± +gd, so v = gd = (9.8)(5) = ± m/s. We choose the egative solutio, because we have defied + to be upward, ad we kow the velocit is dowward. So ow we have two equatios: = cosφ v = ω siφ with two ukows, ad φ = ( ωt + φ). Let s combie the equatios to elimiate φ ad solve for. Use the fact that cosφ = si φ : = si φ v where, from the velocit equatio, we have siφ =. Pluggig this i, we have: ω v = ω v ω = Solvig for : = v ω v = ± + = ±.5 + = ±.75 m ω Choose the positive solutio sice is positive. The maximum legth of the bugee cord will be the equilibrium legth plus the amplitude of S.H.M.: L max =L += = 5 m. This should have two sigificat figures, so studets ma either aswer 5 m or 5. m. L max = 5 m Part D ( poits = 4 poits for each solutio) mplitude The amplitude of S.H.M. will be the maximum legth of the bugee foud from Part C mius the equilibrium legth foud from Part : = L max L = =.75 m. This should have two sigificat figures, so we should report fiall that =3 m. gular Frequec k 68 The agular frequec will be ω = = = radias per secod. This should have two sigificat m 85 figures, so we should report fiall that ω=.89 rad/s. Page 7 of 8

8 PHY 38Y Waves Test Detailed Solutios 5 Phase Costat Whe t=, the legth of the bugee is 5 m. Its equilibrium at = is whe L=L =7.5 m, so the value of at t= is =.5 m. We also kow the velocit is dowward, or egative at this istat. = cos( ωt + φ) = cosφ φ = cos = cos.5.75 φ = ±. radias. The velocit at t=, v = ω siφ must be egative, therefore siφ must be positive, so ol the +. positive solutio is possible. This should have two sigificat figures, so we should report fiall that φ =. rad. Note that a phase agle that is differet tha this value b N(π) is also a correct solutio. For examples, φ =.6, 5.3, 7.3 or 3.6 radias are all also correct. = 3 m ω =.89 rad/s φ =. rad (or. + N( π) where N=a iteger) Page 8 of 8