Physics Spring 2010 Final Exam Solution

Transcription

1 Physics Spring 2010 Final Exam Solution Grading note: There are 12 problems. Point values are given with each problem. They add up to 155 points. In multi-part problems, points are not necessarily evenly divided between the parts. 1. [10 points] A person hums into the top of a well and finds that standing waves are established at frequencies of 42, 70, and 98 Hz, but no frequencies in between these. The frequency of 42 Hz is not necessarily the fundamental frequency. The speed of sound is 343 m. How deep is the well? s A well is open on the top and closed on the bottom. The fundamental frequency of sounds waves in an open-closed tube is f 1 = v 4L, where v is the speed of sound and L is the length of the tube (the depth of the well). Other resonant frequencies are f m = mf 1, where m is an odd integer, that is, they are f 1, 3f 1, 5f 1, 7f 1,.... Notice that each resonant frequency is 2f 1 higher than the previous one. We need to match the list of frequencies given in the problem to this pattern. Notice that each frequency in the list given in the problem is 28 Hz higher than the previous one: 42 Hz+28 Hz=70 Hz; 70 Hz+28 Hz=98 Hz. Thus 2f 1 = 28 Hz f 1 = 14 Hz From this value we can find the distance: f 1 = v 4L L = v = 343 m s 4f 1 (4)(14 Hz) = 6.1 m. Another way to calculate this is to say that if one of the frequencies (42 Hz, say) is the m th harmonic, then the next highest frequency (70 Hz) must be the (m+2) th harmonic: 42 Hz = mf 1 70 Hz = (m + 2)f 1. Subtracting the upper equation from the lower equation gives 70 Hz 42 Hz = (m + 2)f 1 mf 1 28 Hz = 2f 1, from which f 1 and L can be found as above. This same analysis works using 70 Hz as the m th harmonic and 98 Hz as the (m+2) th harmonic. 1

2 2. [10 points] Two pulses travel down a string in opposite directions at speed 1 cm s. The pulses are each half up and half down, as shown in the figure. The figure shows the positions of the pulses at time t = 0. Draw the shape of the string at t = 1 s, 2 s, 3 s, and 4 s. 1 cm/s 1 cm/s To draw the shape at any given time, t, shift the left pulse rightward by t centimeters, shift the right pulse leftward by t centimeters, as shown in the plots on the left below. Then add the two waveforms together, as shown on the right. Notice that at t = 2 the two pulses exactly cancel out. t=0s t=0s t=1s t=1s t=2s t=2s t=3s t=3s t=4s t=4s 2

3 3. [15 points] For each of the two situations given below, give (i) the type of image formed (real or virtual), (ii) whether the image is upright or inverted, (iii) how tall the image is, and (iv) where the image is located (say whether it is to the right or the left of the lens, and how far from the lens it is located). (a) A 20 cm tall flower is placed 30 cm to the left of a converging lens of focal length 10 cm. f = 10 cm, s = 30 cm, h = 20 cm. 1 s + 1 s = 1 f 1 30 cm + 1 s = 1 10 cm s = 15 cm M = h h = s s h cm = cm 30 cm h = 10 cm (i) Positive s implies real image. (ii) Negative y implies inverted image. (iii) Size of y is 10 cm.. (iv) Positive s gives location of image as the opposite side of the lens as the object, that is, 15 cm to the right of the lens. (b) A 20 cm tall flower is placed 30 cm to the left of a diverging lens of focal length 10 cm. f = 10 cm, s 0 = 30 cm, y 0 = 20 cm. 1 s + 1 s = 1 f 1 30 cm + 1 s = 1 10 cm s = 7.5 cm M = y y = s s y cm = cm 30 cm y = 5 cm (i) Negative s implies virtual image. (ii) Positive y implies upright image. (iii) Size of y is 5 cm.. (iv) Negative s gives location of image as the same side of the lens as the object, that is, 7.5 cm to the left of the lens. 3

4 4. [10 points] A prism is made of glass with index of refraction The angles as of the prism are shown in the diagram. A beam of light enters the prism perpendicular to one side as shown in the diagram. At what angle, relative to the horizontal, does the beam of light leave the prism? 30 o incoming light beam o o θ =30 o 1 θ 2 90 o 60 o See the annotated diagram above. Coming into the prism, the beam is perpendicular to the surface and parallel to the normal line. Its incoming angle is therefore θ = 0 (not explicitly shown above). Plug sin 0 = 0 for the incoming angle into Snell s law to find that the outgoing angle is also 0. In other words, because the beam is perpendicular to the first surface, it does not bend there. When the beam hits the far side of the prism, the angle relative to the normal line is 30. (See diagram above. The angle of 30 may be found by considering the angles in the triangle above the light beam in the diagram.) Use Snell s law, with n 1 = 1.50, n 2 = 1.00, and θ 1 = 30 to find θ 2 : n 1 sin θ 1 = n 2 sin θ 2 sin θ 2 = n 1 sin θ 1 = 1.5 n sin 30 = 0.75 θ 2 = 48.6 Angle θ 2 is relative to the normal line. The problem asks for the angle relative to the horizontal, which is 30 less than θ 2, that is, See diagram below =18.6 4

5 5. [20 points] A velocity selector is used in experiments where all the charged particles in a beam are required to have the same velocity (for example, when entering a mass spectrometer). A velocity selector has a region of uniform electric and magnetic fields that are perpendicular to each other and perpendicular to the motion of the charged particles. The electric field and magnetic fields each exert a force on the particles. If a particle has precisely the right velocity, the two forces exactly cancel and the particle is not deflected. Equating the electric and magnetic forces gives the following equation, which we can solve for velocity: qe = qvb v = E B A particle moving at this velocity will pass through the region of uniform fields with no deflection, as shown in the figure. For higher or lower velocities than this, the particles will feel a net force and will be deflected. A slit at the end of the region allows only the particles with the correct velocity to pass. The figure shows E down parallel to the page, and B going into the page. A particle of mass m and charge q follows the path marked v. B particle source v E (a) Assuming that the particle in the figure is positively charged, what is the direction of the force due to the electric field? From F = q E, the direction of the force is the same as the direction of the electric field for a positively charged particle. In this situation, it is down. (b) Assuming that the particle in the figure is positively charged, what is the direction of the force due to the magnetic field? One way to solve it. The fields have been arranged so that the magnetic force is opposite the electric force, so it must be up. Another way to solve it. From F = q v B, use the right hand rule to find the direction of the magnetic field. For v to the right and B into the page, F is up. This problem is continued on the next page... 5

6 (c) How does the kinetic energy of the particle shown in the figure change as it traverses the velocity selector: does it increase, decrease, or stay the same? There is no net force on the particle. Its speed does not change, and its kinetic energy stays the same. (d) Suppose a particle with twice the velocity of the particle in the figure enters the velocity selector. How will the path of the particle curve: upward (toward the top of the page), downward (toward the bottom of the page), out of the page, into the page, or not at all? Magnetic force: F = q v B. Doubling v doubles the (upward) magnetic force. Electric force: F = q E. Doubling v does not change the (downward) electric force. Since the upward force is increased while the downward force in unchanged, the path of the particle will curve upward. (e) Suppose a particle with the same mass and velocity of the particle in the figure but twice as much charge, 2q, enters the velocity selector. How will the path of the particle curve: upward (toward the top of the page), downward (toward the bottom of the page), out of the page, into the page, or not at all? Both the electric and magnetic forces are proportional to charge. Doubling the charge would double each force, so the two forces would still cancel out. The particle would curve not at all. 6

7 6. [10 points] A resistor has a conical shape, as shown in the figure. The top of the resistor has a smaller radius than the bottom of the resistor. It is hooked to a 10 V battery, so that there is a 10 V drop from the top of the resistor to the bottom. Consider the voltage difference between the middle of the resistor (halfway between top and bottom) and the bottom. Is this difference greater than, equal to, or less than, 5 V? Briefly explain your reasoning. V=? Think of cutting the resistor into two parts at the midpoint, as suggested by the figure below, so that each part has the same length, L. Because the cross-sectional area A of each part varies over the length of the part, we cannot precisely calculate the resistance of each part using R = ρl A. However, we can make the qualitative statement that A throughout the lower part is greater than A throughout the upper part. Thus the resistance of the lower part must be smaller than that of the upper part. The two parts are in series, they have the came current. From V = IR, because the lower part has less resistance, it has less voltage. The two voltages must add up to 10 V. In order for the lower part to have less voltage, it must be less than 5V. R bigger R smaller 7

8 7. [20 points] Two resistors, two batteries, two switches, and a capacitor are wired up as shown. Initially the switches are open (as shown) and the capacitor has no charge. R1=5Ω Switch 1 Switch 2 R2=5Ω 5V C=0.1F (a) At time t = 0, both switches are closed. Give a formula for the voltage across the capacitor as a function of time, V C (t), and give a rough sketch this function. Your formula should be a numerical expression with t as the only variable. 5V R1=5Ω R2=5Ω C=0.1F 5V 5V See the redrawn circuit at left. There is always a 5 V difference across the series combination of resistor R2 and the capacitor, since this combination is parallel to the 5 V battery. The presence of the 10 V battery and R1 do not affect the behavior of R2 and C because their total voltage is fixed at 5 V. Thus this situation is exactly the same as a capacitor charging up through a resistor attached to a single 5 V battery. The final voltage is the battery voltage, V f = 5 V. The time constant is τ = (R2)(C) = 0.5 s. Thus V C (t) = 5 V ( 1 e t/0.5 s). (b) If resistors R1 and R2 are light bulbs, describe qualitatively their behavior over time, starting at t = 0. See the diagram in the answer to part (a). Resistor R1 always has 5 V across it, so it is steadily illuminated. Resistor R2 starts off with 5 V (because the capacitor is initially uncharged), but its voltage drops off as the capacitor is charged up. Thus they start off equally bright; R1 does not change; R2 fades out with time constant 0.5 s. This problem is continued on the next page... 8

9 (c) At some much later time, switch 2 is opened. Switch 1 remains closed. Give a rough sketch of the voltage of the capacitor over time, V C (t), starting when switch 2 is opened. What is the characteristic time scale on which the capacitor changes? (You are not required to give a formula for V C (t). Just draw the sketch and calculate the characteristic time scale.) See the circuit drawing at the bottom of the page. Once the switch is opened, current cannot flow through the 5 V battery, so it has no effect on the behavior of the circuit. In effect, the circuit is a 10 V battery with a series combination of R1 and R2 (whose equivalent resistance is 5Ω+5Ω=10Ω) and the capacitor. Before the switch was opened, the capacitor voltage was 5 V. (This was the final voltage in part (a).) After it is opened, the capacitor charges up, asymptotically approaching 10 V. The time scale is τ =RC =(10)(0.1)= 1 s. (d) If resistors R1 and R2 are light bulbs, describe qualitatively their behavior over time. Specifically, state what changes (if any) happen at the moment switch 2 is opened, and describe what (if anything) happens after that. Before the switch was opened, as described in (b), bulb R1 has 5 V and bulb R2 has 0 V; thus R1 is lit up and R2 is dim. Immediately upon opening, the 10 V battery voltage is divided between the capacitor and the two resistors. The capacitor gets 5 V (see part (c)), so the remaining 5 V is divided between the resistors, that is, they each get 2.5 V. This makes them dimmer than R1 was before the switch was opened. As the capacitor charges up, the voltage across the bulbs falls off with time constant 1 s, as calculated in (c). Writing in terms of brightness (power), we have the following. When switch 2 is opened, bulb 1, which had been bright, gets dimmer; bulb 2, which had been off, turns on with the same intensity as bulb 1. They both then fade out with time constant 1 s. R1=5Ω R1=5Ω Switch 2 R2=5Ω R2=5Ω 5V C=0.1F C=0.1F 9

10 8. [15 points] Asteroids are small bodies, typically a few kilometers in size, orbiting the Sun. In 1993, the space probe Galileo flew near asteroid 243 Ida and discovered that it was orbited by a moon. The moon was named Dactyl. The orbit of Dactyl isn t precisely known. For this problem, assume that it moves around Ida with speed 10 m at a distance 90 km from the center of Ida. s Assume that both Dactyl and Ida are bright (i.e., they reflect enough light to be seen) and that they both can readily reflect radar signals. (a) The Hubble space telescope has diameter 2.4 m and uses optical light. (Take λ = 500 nm for this problem.) How close to the Earth would 243 Ida have to be in order for Hubble space telescope to resolve the 243 Ida and Dactyl (i.e., to tell that they are two separate objects)? Telescope resolution (small angle formula, θ in radians): θ λ D Geometry (see figure; small angle formula, θ in radians): θ tan θ = x d λ Equate these formulas: D = x d d = xd λ = ( m)(2.4 m) ( m) = m. Depending on the moon s position in its orbit, its apparent separation when observed from Earth might be less than 90 km (for example, if the moon were in front of the asteroid rather than next to it), in which case smaller distance would be needed. Dactyl x 243 Ida d θ Hubble (b) Imagine a space probe with a radar transmitter and receiver located near the asteroid and moon. At some moment, the asteroid and moon are approaching the space probe, with the asteroid having speed 100 m relative s to the probe. The space probe emits a radar signal toward the asteroid and moon with frequency GHz. Describe the reflected signal detected by the space probe. Depending on how you approach this, your answer may be a mix of quantitative and qualitative parts. First consider the asteroid. It is moving at 100 m s toward the wave source (the space probe), so v r =100 m s. The wave speed is v =c= m s, and the source frequency is f s = Hz. Plug these numbers in to the Doppler formula for reflected waves: f = 2 v ( ) r 100 m v f s s = (2)( m )( GHz) = GHz = 5300 Hz. s The reflector (asteroid) is moving toward the source, so the reflected wave is observed at a higher frequency, f o = f s + f = GHz. What about the moon? It is moving at 10 m s relative to the asteroid, but we don t know the direction. It could be moving as fast as 110 m s toward us or as slowly as 90 m s toward us. Plug this range of velocities into the reflected Doppler equation to find f between GHz and GHz. Thus the observed signal is between GHz and GHz. Summary. A signal at GHz is detected from the asteroid. A signal between GHz and GHz is detected from the moon. 10

11 9. [10 points] A student studying the motion of dancers has a device which measures forces the dancers feet impose on the floor. It produces a signal whose voltage is proportional to the force. The motions made by the dancers which she is interested in measuring take place over time scales of 0.1 seconds and longer. Unfortunately, in addition to detecting the dancers motions, the device vibrates on its own at a frequency of 100 Hz; she does not want this 100 Hz signal corrupting her measurements of the dancers. She wants to use a resistor and a capacitor in a circuit to remove the 100 Hz vibration signal while retaining signals from the dancers movements. (Note: this problem may have a range of possible answers.) (a) Which should she make: a low-pass filter or a high-pass filter? (b) She will use a 10 kω resistor. What value of capacitor could she use? (c) Sketch a diagram of what her circuit should look like. Indicate the input (the signal from the force measurement device) as V in. Indicate the output (which she will record with a computer) as V out. She wants to detect signals with periods of 0.1 s or longer, which corresponds to frequencies of f = 1/T = 10 Hz or smaller. She wants to eliminate signals at 100 Hz. Thus she wants a (a) low pass filter. The characteristic frequency of the filter should be between 10 Hz and 100 Hz. Anything in between is acceptable. We ll pick 30 Hz. (We haven t discussed this in class, but usually one picks an in-between frequency using a logarithmic scale rather than a geometric scale. That puts the half-way-in-between frequency at 30 Hz rather than 55 Hz.) This gives characteristic time scale τ = 1/(30 Hz) = s. Solving τ =RC for capacitance gives C =τ/r=0.033/10000 = F=3.3 µf. In a lower-pass RC filter, the output voltage is across the capacitor, as in the drawing below. The order of the resistor and capacitor doesn t matter; what s important is that V out is parallel to the capacitor. R V in C V out 11

12 10. [10 points] An infinitely long, straight wire carries a current, I 1, which steadily increases, going from 0 to 6 Amp in 0.2 seconds. A conducting coil is 0.1 m away from the wire. The coil consists of 1000 turns with radius 0.01 m. It is oriented as shown, with the coils parallel to the plane of the paper. The ends of the coil are attached to a resistor of resistance R = 1Ω. What is the current through the resistor? Give both the magnitude of the current and its direction, up (toward the top of the page) or down (toward the bottom of the page). Note. Because the radius of the coil is much smaller than its distance from the wire, you may assume that, at any instant, the magnetic field is the same everywhere in and near the coil. I 1 R 0.1 m Direction. According to the right hand rule, the upward current in the wire, I 1, produces a magnetic field into the page at the location of the coil. The current, I 1, is increasing, so the magnetic field through the coil, and hence its flux, into the page, is increasing. A current is induced in the coil to oppose this field, so the induced current will produce a field within the coil which points out of the page. According to the right hand rule, this requires a counterclockwise current around the coil. Following the geometry of this wire, such a current would flow up, toward the top of the page when going through the resistor. Magnitude. Let d be the distance from the wire to the coil and r be the radius of the coil. The area of the coil is A = πr 2. B = µ 0I 1 2πd Φ B = BA = µ 0I 1 πr 2 = µ 0I 1 r 2 2πd 2d E = N Φ t = Nµ 0r 2 I 1 2d t I = E R = Nµ 0r 2 I 1 2dR t = (1000)(4π 10 7 )(0.01) 2 (6) 2(0.1)(1)(0.2) = Amp 12

13 11. [10 points] (a) You have a rectangular block of dielectric material with dimensions 10 cm 20 cm 40 cm. You can put metal plates on any two opposite sides of the block to form a capacitor. Which sides of the block should you put the plates on to maximize the capacitance? (Answer A, B, or C, based on the figure; obviously, one of the plates will be on a side of the block not visible on the figure.) 10 cm 40 cm A C B 20 cm Use C = ɛa d. To maximize C, you need to maximize A while minimizing d. This is done by putting the plates on the 20 cm 40 cm sides, i.e., side C. (b) Consider the two capacitors shown below. The first one, in the upper figure, has plate separation d. The second one, in the lower figure, has plate separation 2d but also has a metal slab of thickness d in the middle. Is the capacitance of the second capacitor greater than, less than, or the same as the capacitance of the first capacitor. conductor? Hint. One way to start this problem is to compare what happens if you put charge Q on each capacitor. 2d d d/2 d d/2 V 1 V 1/2 V 1/2 Consider charge Q on the original capacitor, as in the figure. This produces electric field E = Q ɛ 0A inside the capacitor. To find the voltage, rewrite E = V x as V 1 =E x=ed= Qd ɛ. 0A Now consider the same charge Q on the second capacitor. Charges Q and +Q are induced on the center conductor. In the gaps between plates and conductors, the electric field is E = Q ɛ 0A, just as the original capacitor. In each gap, the distance is half the d original distance,, so the voltage is half the original Q d 2 ɛ 0A = Qd 2 voltage: 2ɛ = V1 0A 2. The two gaps are in series, V so their voltages add up, V1 2 = V 1. Thus the voltage difference between the top and bottom plate is the same as in the original capacitor. Capacitance is defined through Q = CV. Since, for a given Q, both capacitors give the same voltage, V 1, they must have the same capacitance. Note: you can think of this as two capacitors in series. However, unlike resistors, you can t just add the capacitances of two capacitors in series. See text for details. 13

14 12. [15 points] A person consumes 2000 calories in a day. This means that he whatever food is needed for his body s metabolic process to generate 2000 calories of energy a day by digesting the food. His mass is 75 kg and the surface area of his body is 2.0 m 2. A food calorie is really a kilocalorie, which is equal to approximately 4186 J. To convert from Celsius to Kelvin: T (Kelvins) = T ( C) Latent heat of fusion of water L f = 334, 000 J kg Latent heat of vaporization of water L v = 2, 260, 000 J kg Specific heat of ice c ice = 2090 J kg C Specific heat of liquid water c water = 4186 J kg C Specific heat of steam (at constant pressure) c steam = 2010 J kg C (a) If he retained all of the food energy in his body, by how much would his temperature increase in one day? (Assume that he is made entirely of water.) Total energy input is Q=(2000 kcal)(4186 J/kcal)= J. Q = mc T T = Q mc = J ( (75 kg) 4186 J kg C ) = 27 C This problem is continued on the next page... 14

15 (b) In order to maintain temperature equilibrium, he must get rid of 2000 calories of energy a day. One way to do this is by sweating, in which the energy is used to evaporate water. A model for this evaporation is that energy from the body is used to bring the water from human skin temperature, 34 C, up to the boiling point of water, 100 C, and then to vaporize the water. (Note. Although water evaporated from human skin is really evaporated at a lower temperature than 100 C, the heat-then-vaporize model suggested here gives reasonably accurate numerical results.) Assume that sweating is the only mechanism by which he loses energy. What mass of water he evaporate every day in order to maintain a steady temperature (in other words, how much must he sweat)? As in part (a), total energy input is Q=(2000 kcal)(4186 J/kcal)= J. The water must be increased from 34 C to 100 C, a difference of T = 66 C, and then vaporized. Q = mc T + ml v Q = m(c T + L v ) m = Q c T + L v = ( J ) 4186 J kg C (66 C) + 2, 260, 000 J kg = J 2, 540, 000 J kg = 3.3 kg 15