Physics Spring 2009 Midterm #1 Solution

Transcription

1 Physics 02- Spring 2009 Midterm # Solution Grading note: There are 7 problems. Each problem is on its own page. Point values are given with each problem. They add up to 70 points. In multi-part problems, points are not necessarily evenly divided between the parts.. [5 points] (a) President Obama recently proposed a dollar budget for 200. What is the in (i) Megadollars? (ii) Teradollars? (iii) Nanodollars? (There is a typo; the problem should read What is this in... ) ( ) megadollar (i) ( dollar) 0 6 = megadollars dollar ( ) teradollar (ii) ( dollar) 0 2 = 3.6 teradollars dollar ( ) nanodollar (iii) ( dollar) 0 9 = nanodollars dollar (b) In conventional mks units, Newton s gravitational constant is G = m3 kg s 2. In some fields of physics, cgs units are used; they are based on units of cm for distance, g for mass, and s for time. What is G in these units? ( ) 3 ( ) m3 cm kg 8 cm kg s m 0 3 = g g s 2 (c) The North Pole is 0,000 km from the Equator, following a path along the surface of the Earth. In fact, that was how the French Academy of Sciences originally defined the meter. In their honor, we ask the following question. Euro coins (e) are mm in diameter. How many such coins, laid next to each other, would it take to go from the Equator to the North Pole? 0000 km mm ( 0 3 ) ( ) m mm km 0 3 = m

2 2. [5 points] Measurements of sound levels at the center of a football stadium find that when the stadium is filled and the entire crowd is shouting, it is 48 db louder than when a single person in the stands is shouting. Assuming that each person in the stands is equidistant from the center of the stadium, and that all people shout equally loudly, how many people does the stadium hold? Let I be the intensity of one person shouting. Then ni is the intensity of n people shouting. 48 db = 0 log ni I 48 = 0 log n 4.8 = log n = 0 log n 63, 000 = n 2

3 3. [0 points] Doppler radar stations track weather systems such as Monday s snowstorm by sending out pulses of radar signals. The radar signals travel at the speed of light. The signals bounce off of water molecules in the atmosphere and arrive back at the radar station. While tracking a certain storm cloud, pulses are sent out at a rate of 0, Hz. The reflected pulses are detected starting s after the initial transmission, and the return signal consists of pulses which come in at a rate of 0, Hz. (a) How far away is the storm cloud? (b) How fast is it moving? Toward the radar station, or away from it? (a) The round-trip time of a pulse is t = s. In this time, traveling at a speed of light, the pulse travels total distance x = vt = ( m )( s) = 50, 000 m. s This is the round-trip distance to the cloud, so the one-way distance is 75,000 m=75 km. (b) The shift in frequency is f = 0, Hz 0, Hz = Hz The signal is reflected off of a moving cloud. This means that it is effectively Doppler shifted twice, once when the cloud is the observer receiving the pulses and once when the cloud is the source, re-transmitting the pulses back to the radar station. We developed the following formula for use in this situation... f = 2 v reflector f s v v reflector = v f 2f ( s ) m s ( Hz) v reflector = (2)(0, 000 Hz) v reflector = 2 m s Because the frequency has gotten higher, we know that the cloud is moving toward the radar station. One way to understand this is that, as the cloud moves toward the radar station, each wave crest has to travel a slightly smaller distance than the previous wave crest. Thus the time between observed wave crests is shorter than the time between emitted wave crests, so the observed frequency is higher. 3

4 4. [0 points] A 000 kg car is at rest. When a 50 kg person gets into it, the car body sinks down by 2 cm as springs in its shock absorbers are compressed. Someone pushes it down 2 cm further and lets go. The car oscillates up and down. (a) What is the period of oscillation? (b) What is the maximum acceleration felt by the person inside the car? (a) First, find the spring constant, k. When the 50 kg person gets into the car, the additional weight of the person is balanced by the compression of the springs, x = 2 cm = 0.02 m: mg = k x k = mg x = (50 kg)(9.8 m s ) = 24, 500 N 0.02 m m. Note: only the additional weight of the 50 kg person is used in this calculation. The shock absorber springs had already been compressed by the weight of the car when the car was sitting at rest at the start of the problem. Next, find the frequency, ω, and period, T. Both the car and the person in it are bouncing up and down, so use the total mass 050 kg when calculating the frequency of oscillation. ω = k m T = 2π ω = 2π m k = 2π 050 kg 24, 500 N m =.3 s (b) The car was pushed down an additional 2 cm in order to get it oscillating, so the amplitude of the oscillatory motion is A = 2 cm. The acceleration as a function of time is a(t) = Aω 2 cos ωt. Since cosine ranges from to +, the extreme values of a(t) are ±Aω 2, and the maximum acceleration is ( ) k Aω 2 = A = 0.46 m m s 2. 4

5 5. [0 points] A laser of wavelength 632 nm shines on a set of diffraction slits. The light emerging from the slits shines on a screen 0 m away, producing the pattern shown below. Larger spots on the figure below represent brighter dots seen on the screen. There are additional spots off either end of the figure which are not shown. The measurements along the ruler in the figure are in millimeters. What are (a) the slit width and (b) the slit spacing? The solid line on this graph shows intensity versus position of the spots. The dashed line is what we infer to be the envelope due to the width of the slits. Notes: Ignore the scruff between the central peaks. In this plot, the low-level peaks have been exaggerated relative to their true values. 400 (a) The central bright spot is at 200 mm on the ruler. Consider the pattern to the right of the central point. (We could equally well consider the pattern to the left, and we would reach the same conclusions.) The spots get dimmer out to a ruler positions of around 270 mm. The n = dark spot for single-slit diffraction must be somewhere around here. Evidently this dark spot doesn t precisely align with a spot from the double-slit pattern, so there is no missing spot. Nevertheless, we could use an estimated value of y = 70 mm and n = to calculate a. As we move further out, the spots get brighter (e.g., at 300 mm) and dimmer again, completely disappearing at ruler position 340 nm. This must be the n = 2 dark spot. Since we know its value precisely, we will use it, rather than the n = dark region, to determine the slit spacing. For this spot, we have y = 340 mm 200 mm = 40 mm = 0.4 m. With n = 2, L = 0 m, and λ = 632 nm = m, we have y = nl λ a a = nlλ = (2)(0 m) ( m ) = m = 90 µm. y 0.4 m (b) The spacing between the bright spots is 20 mm=0.02 m. separated by: Thus the slits are y = nl λ d d = nlλ = ()(0 m) ( m ) = m = 320 µm. y 0.02 m 5

6 6. [0 points] An object,.00 cm tall, is placed 2.0 cm to the left of a diverging lens, f =-8.00 cm. A converging lens is placed 8.00 cm to the right of the diverging lens. The final image is virtual and is 29.0 cm to the left of the diverging lens. IMAGE Not to Scale OBJECT Image formed by First Lens s 2 s s f = 8 cm 2 f =? cm x(cm) (a) What is the focal length of the converging lens? (b) What is the height of the final image? The figure above has been modified for the exam solution. It shows the image formed by the first lens, along with its distance from the first lens, s, and from the second lens, s 2. The figure also shows the final image distance, s 2. (a) First lens: The object is s =2 cm to the left of the lens. + s s = f 2 + s = 8 s = 4.8 cm This is a virtual image, 4.8 cm to the left of the first lens, which puts it at position x = 7.2 cm on the ruler in the figure. Second lens: See the figure. The first image is s 2 =2.8 cm to the left of the second lens. The final image is s 2= 37 cm to the left of the second lens; a negative number is used because it is a virtual image. + s 2 s = 2 f = f 2 = 9.6 cm. f 2 (b) Lens : M = s s = = Lens 2: M 2 = s 2 s 2 = = Total: M = M M 2 = (0.40)(2.89) =.2 Since M = h /h, we have h = Mh = (.2)( cm) =.2 cm. 6

7 7. [0 points] There is a hoop of radius R. The hoop is made of flexible material, like a slinky, so that waves can travel around it. The waves travel at speed v. A periodic wave traveling around the hoop will catch up with itself and interfere with itself. Under the right circumstances, this results in a stable (resonant) pattern on the hoop. At what frequencies will such stable patterns arise? Your answer should be a mathematical expression which depends on R and v. Hint: first find wavelengths, then find frequencies. By the way, this problem is closely related to the quantum mechanical model of electrons orbiting atomic nucleus. v R Consider the interference between a wave traveling through some point on the ring and the next time the wave travels through that same point on the ring. The second time, it has gone an extra path length, L = 2πR. Our interference condition is that we get constructive interference (giving a stable or resonant pattern) when L = nλ for any integer n. Thus the wavelengths at which constructive interference occurs are The frequencies are 2πR = nλ λ = 2πR n v = fλ f = v λ = n v 2πR. Another way to think about this is that wave pulses going around the ring are like wave pulses on a stretched slinky making a round trip from one end to the other and back again. The total length of that round trip on a slinky has to be an integer number of wavelengths, nλ. For a slinky, since the round-trip distance is 2L, this meant that 2L = nλ, so λ = 2L/n and f = v/λ = nv/2l. For the hoop, the distance is 2πR instead of 2L. 7