Physics 102-1 Spring 2008 Midterm #1 Solution Grading note: There are seven problems on nine pages. Point values are given with each problem. They add up to 110 points. In multi-part problems, points are not necessarily evenly divided between the parts. 1. [15 points] (a) What does the metric prefix Tera mean? 10 12 (b) Gravitational acceleration g is 9.80 m km. What is this in units of s2 µs? 2 ( 9.80 m ) ( ) ( 1 km 10 6 ) 2 s 15 km s 2 10 3 = 9.80 10 m 1 µs µs 2 (c) A computer clock runs at a rate of 2.0 GHz. What is the period of this clock in seconds? What is the period in ns? T = 1 f = 1 2.0 GHz ( ) 1 GHz 10 9 = 5 10 10 s Hz ( ) 1 ns 5 10 10 s 10 9 = 0.50 ns s 1
2. [15 points] (a) Which sounds louder, 1 frog 1 m away, or 25 frogs 10 m away? Assume that all frogs produce identical noises. Find the intensity in each case in terms of the (unknown) power emitted by a single frog, P frog : ( ) P frog 1 Single frog: I s = (1) 4π(1 m) 2 = 4π m 2 P frog ( ) P frog 1 Group of frogs: I g = (25) 4π(10 m) 2 = 16π m 2 P frog = 1 4 I s The intensity of sound from the group is smaller by a factor of 4, i.e., the single frog is louder. (b) How much louder, in db, is the louder group compared to the quieter group? 10 log ( Is I g ) = 10 log ( Pfrog 4π (1 m) 2 25 P frog 4π (10 m) 2 ) = 10 log ( 1 25 100 ) = 10 log ( ) 100 25 = 10 log(4) = 10 (0.60) = 6.0 db 2
Sun Diffraction Grating Wall of Space Station A B C D 3. [10 points] An astronaut on the space station wants to get a suntan. She knows that, in addition to visible light, the sun produces ultraviolet light, and that exposing herself to ultraviolet light will give her a suntan. She replaced one of the windows of the space station with a diffraction grating. Sunlight shining into the window gets diffracted onto the wall opposite the window, with different colors of light ending up at different points on the wall. The first order spectra are spread out over the two locations indicated by long gray rectangles in the diagram. Of course, the real spectra would have a rainbow of colors, going in one direction or the other from red to violet, but the colors are not shown in the figure. The astronaut knows that the wavelengths of ultraviolet light are shorter than the wavelengths of visible light. At which of the position(s), A, B, C, and/or D, indicated in the figure should she locate herself to get a suntan? Bright spots in a diffraction pattern are at positions y = nl λ d, where y is the distance from the center of the pattern. The important point here is that y is proportional to wavelength λ. The wavelength of ultraviolet light is shorter than that of visible light, so the ultraviolet light will be at relatively small distances from the center of the pattern. Thus the astronaut should go to point B or C to get her suntan. 3
4. [20 points] The figure shows two identical springs, each with spring constant 120 N/m, one attached to mass m 1 and the other attached to mass m 2. The springs are initially stretched out as shown, with m 1 pulled out 0.10 m from equilibrium and m 2 pulled out 0.20 m from equilibrium. The masses are released from rest at these positions at time t = 0. 0.10 m m1 m2 0.20 m (a) How much time passes between the moment the masses are released when they are side by side for the first time at x = 0? Assume the masses are identical, with m 1 = 3.0 kg and m 2 = 3.0 kg. See the solution on the following page. (b) How much time passes between the moment the masses are released when they are side by side for the first time at x = 0? This time, assume the masses are m 1 = 3.0 kg and m 2 = 27.0 kg. See the solution on the page after next. This problem is continued on the next page... 4
Solution to (a) The masses and the spring constants are identical, so the two masses oscillate with the same angular frequency, ω 1 = k/m = 6.32 s 1, and hence have the same period, T 1 = 2π/ω = 1.0 s. (The amplitudes of the masses are different, but this does not affect their periods.) After one quarter of a period, both of the masses will pass through the equilibrium point, x = 0 (see graph below). This is what the problem is asking about, so the answer is T 1 /4 = 0.25 s. Note: the equations for the positions of the two masses are x 1 (t) = 0.10 cos(ω 1 t) x 2 (t) = 0.20 cos(ω 1 t) So the time at which they pass could be solved by equating these expressions: 0.10 cos(ω 1 t) = 0.20 cos(ω 1 t). This sort of equation is, in general, very difficult to solve for t. In this particular problem, you are told that the masses pass each other at x = 0, so you could set either x 1 (t) or x 2 (t) to zero and deduce that cos(ω 1 t)=0. The cosine function is zero when its argument is π/2, 3π/2, etc. We are interested in the first time it crosses zero, so we have ω 1 t = π/2. Inserting ω 1 = 6.32 s 1 and solving for t gives t = 0.25. Although this line of reasoning works, it is easier to think the problem through physically (rather than mathematically) as done in the first paragraph of this solution. 4a
Solution to (b) Mass 1 behaves is in part (a), but mass 2 now moves more slowly, with angular frequency ω 2 = k/m 2 = 2.10 s 1 and period T 2 = 2π/ω = 3.0 s. The important thing here is that the period of mass 2 is three times the period of mass 1. This results directly from the facts that m 2 = 9m 1 and 9 = 3. So, what happens when they are released? See the plot below. After 0.25 seconds, mass 1 goes through the equilibrium point, but mass 2 has not yet reached it. After 0.50 seconds, mass 1 is fully to the left, but mass 2 has still not yet reached the equilibrium point. Finally, after 0.75 s, which is (3/4)T 1 and (1/4)T 2, mass 1 is once again going through the equilibrium point, moving to the right, and mass 2 has finally reached the equilibrium point, moving left. In other words, both masses are at x = 0, so this is the point we are looking for. As in part (a), one could, in principle, solve this by writing out the equations for x 1 (t) and x 2 (t), equating them, and solving for t. However this mathematical approach is much more complicated then the physical reasoning outlined above. 4b
(c) Go back to the initial assumption that the masses identical, m 1 = 3.0 kg and m 2 = 3.0 kg, and that they are released at the positions described above. Each of the masses has, attached to it, a small speaker which emits sound waves at 500 Hz. You are standing off to the right. As the masses move back and forth on the springs: i. What is the highest frequency sound you hear? ii. Which mass does this highest frequency come from (m 1, or m 2, or both)? iii. When do you hear this highest frequency? (Either give the amount of time after which the masses were released or give a description of the position and direction of motion of the mass.) Assume the sound waves travel at 343 m s. You hear variations in frequency because of the Doppler effect. You (the observer) are not moving, but the sources (the speaker attached to the masses) are. Putting v 0 = 0 into the Doppler equation gives: 1 f o = (1 vs v )f s. The observed frequency, f 0, will be highest when the denominator has a negative sign. (This makes the denominator smaller than 1, which makes f o larger.) That means the frequency is higher when the source is coming toward you when its speed v s is largest. For either mass, the speed is largest (iii) when it is moving through the equilibrium point to the right. The first time this happens is at t = (3/4)T = 0.75 s, and it happens again once per period (i.e., once a second). The speed of either mass can be written v(t) = Aω sin(ωt). Since sin(ωt) must range between 1 and 1, the maximum value that v(t) can have is Aω. Since mass 2 starts farther out than mass 1, mass 2 has the larger amplitude, A = 0.20 m, so (ii) mass 2 has the higher maximum velocity, v = Aω = (0.20 m) ( 6.28 s 1) = 1.26 m s. Plug this into the Doppler formula to get 1 f o = (1 1.26 (500 Hz) = (i) 501.8 Hz m s 343 ) m s 5
5. [20 points] Light of wavelength 500 nm in air shines onto a pair of slits. The slits are each 100 µm wide, and the centers of the two slits are 300 µm apart. (a) Sketch a graph of intensity versus position which would appear on a wall 2.0 m away from the slits. Your graph should include enough information (axis labels, etc.) so that all the important physical features and their dimensions are evident. (Hint: The angles in this problem are small.) The outer envelope of the pattern, due to the width of the individual slits, has a central bright region surrounded by much weaker bright regions. Dark points between the bright regions are at y = nl λ a = n(0.0100 m) = n(1.00 cm), n 0. The pair of slits result in alternating bright and dark spots, with bright peaks at y = nl λ d = n(0.0033 m) = n(0.33 cm). The overall pattern looks like this: (b) The space between the light source and the slits continues to remain empty, but now the space in between the slits and the wall is filled with glass with index of refraction 1.500. Does the pattern on the wall differ from that of part (a)? If so, describe how the pattern would differ, either by sketching the new pattern or by giving a succinct, quantitative description of how the pattern would differ from that of part (a). The important thing in two-slit and multi-slit interference is the difference in path lengths of different rays after they passed through the slits. These differences were always written in terms of wavelengths. If the space is filled with material of index of refraction 1.500, the wavelength is shorted to λ = 500 nm/1.500 = 333 nm. The new wavelength is 2/3 its original value. The positions, y, of bright and dark spots are proportional to λ (since y = nl λ a and y = nl λ d ). Since the wavelength is now only 2/3 its original value, the pattern is compressed by a factor of 2/3. The picture looks like this: This problem is continued on the next page... 6
(c) Now, assume that the space between the slits and the wall is empty (as in part a), but the space between the light source and the slits is filled with glass with index of refraction 1.500. (The light source still emits light with the same frequency as before.) Does the pattern on the wall differ from that of part (a)? If so, describe how the pattern would differ, either by sketching the new pattern or by giving a succinct, quantitative description of how the pattern would differ from that of part (a). In two-slit and multi-slit interference, we have always assumed that the light has the same path length to each of the slits. Since there is no L on this side of the slits, the wavelength does not matter. This is different from the situation in (b), where wavelength matters. The pattern is the same as in part (a). (d) Now imagine that we re back to the situation in part (a), with empty space between the light source and the slits, and empty space between the slits and the wall, except that a microscopic piece of glass is placed in front of one of the slits so that a ray of light goes through the piece of glass immediately before going through the slit. The piece of glass is designed so that the light going through the slit is delayed by exactly half of a period compared to light going through empty space, so that light which emerges through this slit is completely out of phase with light emerging from the other slit. How does this change the pattern on the wall, if at all? Either give a sketch, much as you did to answer part (a), or give a succinct description of how this pattern would differ from the one you sketched in part (a). The pattern is reversed: dark spots become bright and vice versa. Why? The light rays coming out of the two slits are out of phase with one another, so if they travel equal path lengths to reach the screen, L = 0, they will cancel out and make a dark spot. The same thing happens for L = nλ where n is any integer. In contrast, if light traveling on one of the paths travels half a wavelength more than the other, the waves will be lined up, making a bright spot. The same things happens for L = ( n + 2) 1 λ with any integer n. Here s what it looks like: 7
6. [15 points] Last Tuesday s New York Times, had an article titled NASA Insists It Can Fix Flaw in Rocket Design. The gist of the article is that the Ares I rocket, designed as a replacement for the Space Shuttle for sending humans into space, might have vibration problems. The first stage of the rocket, the solid rocket motor, is a cylinder 50 m long and 4 m in diameter. The cylinder is initially filled with solid rocket fuel, but the fuel burns up in order to propel the rocket. After the fuel is burned up, the cylinder has some gas in it, presumably left over from the combustion of the fuel. The article says that the cylinder can resonate like an organ tube. The concern is and that the resulting vibrations will be harmful to astronauts riding higher up in the rocket. The Times article says that the resonance frequency is 15 Hz. The article included the following figure to explain the physics of the rocket. (a) If the resonant pattern is the one shown in the diagram from the Times article, what is the wavelength of this mode? The picture shows 3/4 of the wave over the length of the tube. The problem gives the length as L = 50 m. Thus: L = 3 4 λ λ = 4 ( ) 4 3 L = (50 m) = 67 m 3 (b) Usually the dominant resonance in a tube is the lowest-frequency resonance (the fundamental mode ). This is different from the mode shown in the figure. What is the wavelength of the fundamental mode be for this cylinder? The lowest mode of an open-closed tube looks like this:. There is only one quarter of a wavelength over the length of the tube, so L = λ/4, which gives λ = 4L = 200 m. (c) Assume that the 15 Hz mode referred to in the Times article is the fundamental mode. What is the speed of waves in the cylinder? (Note: the cylinder is not filled with air at 20, so the speed is not 343 m s.) v = fλ = (15 Hz)(200 m) = 3000 m s 8
7. [15 points] A wave has amplitude 0.002 m and wavelength 0.5 m. It moves to the right at 125 m/s. (a) What are the period and the frequency of this wave? v = fλ f = v λ = 125 m s 0.5 m = 250 Hz T = 1 f = 1 250 Hz = 0.004 s (b) Write an equation, y(x, t), for the wave. Your answer should be written in terms of t, x, trigonometric function(s), and numbers. There may be more than one acceptable answer. ( 2π f(x, t) = a sin λ x 2π = (0.002 m) cos ) T t ( 2π 0.5 m x 2π ) 0.004 s t = (0.002 m) cos (12.57 x 1570 t) (c) Make two sketches of the wave, one at time t=0 and one at time t=1 ms. The axes in your sketches should be properly labeled in order to convey all the essential properties of the wave. You can draw the two sketches on a single plot or on two separate plots 9