Physics 9 Fall 009 Homework 1 - s 1. Chapter 1 - Exercise 8. The figure shows a standing wave that is oscillating at frequency f 0. (a) How many antinodes will there be if the frequency is doubled to f 0? Explain. (b) If the tension in the string is increased by a factor of four, for what frequency, in terms of f 0, will the string continue to oscillate as a standing wave with three antinodes? (a) Because the ends of the string are tied down, increasing the oscillation frequency will decrease the wavelength. Doubling the frequency will cut the wavelength in half, and so will double the number of antinodes to six. T (b) The velocity of the wave depends on the square root of the tension, v =. µ So, increasing the tension by a factor of 4 doubles the velocity. Since the velocity of the wave is v = λf, if we double v, but want to keep the wavelength fixed (this fixes the number of antinodes), then we would need to double the frequency, which we can see as follows: v = λf 0 v = λ (f 0 ) = λf. So, the new frequency f = f 0. 1
. Chapter 1 - Exercise 18. The lowest note on a grand piano has a frequency of 7.5 Hz. The entire string is.00 m long and has a mass of 400 g. The vibrating section of the string is 1.90 m long. What tension is needed to tune this string properly? The lowest note on the grand piano is the fundamental harmonic, meaning that the wavelength is twice the length of the vibrating section of the string, λ = L, where L = 1.90 m. The speed of the wave is v = λf, but also depends on the tension of the wire as v = T µ, where µ = m/l is the linear mass density of the wire (l is the full length of the wire, not the vibrating part). So, solving for the tension gives With numbers we find T = µ (λf) = 4µL f = 4 ml f. l T = 4 ml f l which is a pretty big amount of force! = 4.400 (1.90) (7.5) = 185 N,
3. Chapter 1 - Exercise 19. A violin string is 30 cm long. It sounds the musical note A (440 Hz) when played without fingering. How far from the end of the string should you place your finger to play the note C (53 Hz)? Since the speed of the wave along the string depends on the tension and mass density of the string, v = T/µ, it is constant. Furthermore, the speed of the wave is related to its frequency and wavelength, v = λf. So, λ 440 f 440 = λ 53 f 53. Since the 440 Hz note is the fundamental frequency, played without fingering, the wavelength λ 440 = L 440 = 30 = 60 cm. The wavelength λ 53 = L 53, and so L 53 = f 440 L 440 = 440 30 = 5. cm. 4 53 53 So, the length of the string is 5. cm. But, the question asks how far from the edge you should put your finger. Since the sting is 30 cm long, but you need it to be 5. cm, you should put your finger back 30 5. = 4.8 cm from the end of the string. 3
4. Chapter 1 - Exercise 4. A very thin oil film (n=1.5) floats on water (n=1.33). What is the thinnest film that produces a strong reflection for green light with a wavelength of 500 nm? The strong reflection occurs when the light reflected from the oil and water interfere constructively. The extra distance that the wave reflected from the water travels is d, where d is the thickness of the oil (it has to go through the oil, then back again). This distance has to be a whole number of wavelengths, mλ. But, in a material of index n, the wavelength is decreased by a factor of n. Here n is the index of the foil, since this is the material through which the light is traveling. So, d = mλ. Since we want the n thinnest film, we ll set m = 1. Then, d = λ n = 500 (1.5) = 00 nm. 4
5. Chapter 1 - Problem 41. Astronauts visiting Planet X have a.5 m-long string whose mass is 5.0 g. They tie the string to a support, stretch it horizontally over a pulley.0 m away, and hang a 1.0 kg mass on the free end. Then the astronauts begin to excite standing waves on the string. Their data show that standing waves exist at frequencies of 64 Hz and 80 Hz, but at no frequencies in between. What is the value of g, the free-fall acceleration on Planet X? The setup is seen in the figure to the right. The speed of thewave depends on the tension in the string, v =, where µ is the mass density of T µ the string. The tension in the string comes from the weight of the 1 kg mass, so T = Mg. So, Mg v =. Again, though, the velocity is v = λf. µ We know two frequencies, but they don t tell us the wavelength for each. Higher harmonic frequencies increase as f m = mf 0, where f 0 is the fundamental frequency, for which λ 0 = L. We know that we have two frequencies, one at 64 Hz, and the next at 80 Hz. From this we can determine the fundamental frequency. Suppose that the 64 Hz frequency is the m th frequency, mf 0 = 64. Then, the (m + 1) th frequency is (m + 1) f 0 = 80. Subtracting the first from the second gives (m + 1) f 0 mf 0 = f 0 = 80 64 = 16. Thus, the fundamental frequency is f 0 = 16 Hz. So, putting it all together gives Mg v = λ 0 f 0 = µ g = λ 0f0 m l M, where we have substituted µ = m/l, where m is the mass of the string, and l is the full length of the string. Finally, the fundamental wavelength is λ 0 = L, and so Inserting values gives g = 4L f 0 l m M. g = 4L f 0 l m M = 4 () (16) 0.005.5 1 which is a fair amount smaller than Earth s gravity. = 8. m/s, 5
6. Chapter 1 - Problem 66. A soap bubble is essentially a very thin film of water (n=1.33) surrounded by air. The colors that you see in soap bubbles are produced by interference, much like the colors of dichroic glass. (a) Derive an expression for the wavelengths λ C for which constructive interference causes a strong reflection from a soap bubble of thickness d. Hint: Think about the reflection phase at both boundaries. (b) What visible wavelengths of light are strongly reflected from a 390-nm-thick soap bubble? What color would such a soap bubble appear to be? (a) The light ray falling on the soap bubble reflects from two places. It reflects from the surface of the bubble, and then it also reflects back from the back of the bubble, after traveling an extra distance x = d (down and back), where d is the thickness of the bubble. The wave reflecting from the top surface reflects from a material having a higher index of refraction. This induces an extra phase shift of π, that the reflection from the back surface doesn t have. The net phase of the first wave, at a fixed time which we take to be t = 0, is Φ 1 = kx 1 + φ 1 + π, where k is the wave number, and φ 1 is the intrinsic phase shift due to the source. Similarly, the net phase of the second wave isφ = kx + φ. The second wave doesn t have an extra phase shift because it s reflecting from the air behind the bubble, which has a lower index of refraction. The phase shifts of these two waves are equal since they both come from the same source, so φ 1 = φ. The difference in waves is Φ = Φ Φ 1 = k (x x 1 ) + (x x 1 ) π = k x π. Now, in order to have constructive interference, the net phase shift has to be a whole multiple of π, and so Φ = k x π = mπ, where m = 0, 1,,. So, if k = π/λ, and x = d, we find λ = d m + 1. Now, here λ is the wavelength in the in the soap, which differs from the wavelength in the air by a factor of n, λ soap = λair. So, we can solve for the wavelengths in n air that we would see λ air = nd m + 1 =.66d m + 1, since n = 1.33. (b) Only m = 1 and gives a reflected wavelength in the visible spectrum. For m = 1, λ = 69 nm, which is reddish, while for m =, λ = 415 nm, which is violet. Together, they look purplish. 6
7. Chapter 1 - Problem 77. Two loudspeakers emit 400 Hz notes. One speaker sits on the ground. The other speaker is in the back of a pickup truck. You hear eight beats per second as the truck drives away from you. What is the truck s speed? Since the truck is driving away from you, it s frequency is Doppler-shifted. The expression for the shifted frequency that we get is f obs = f source 1 + v s /v, where f source is the source frequency, v s is the velocity of the source, and v = 343 m/s is the speed of sound. The beat frequency is the difference between the two frequencies, f beat = f source f obs. So, plugging in for the source frequency to the Doppler formula gives Solving for the velocity of the source gives f obs = f obs f beat 1 + v s /v, v s = f beat v = 8 343 7 /m/s. f source 400 So, the truck is driving away from us at about 7 m/s, or about 16 mph. 7
8. Chapter 1 - Problem 78. (a) The frequency of a standing wave on a string is f when the string s tension is T. If the tension is changed by the small amount T, without changing the length, show that the frequency changes by an amount f such that f f = 1 T T. (b) Two identical strings vibrate at 500 Hz when stretched with the same tension. What percentage increase in the tension of one of the strings will cause five beats per second when both strings vibrate simultaneously? T (a) The speed of a wave on a string is v =. Since v = λf, then f = 1 T. Now, µ λ µ taking the derivative, df dt gives df dt = d dt ( ) 1 λ T µ = 1 1 µt Now, divide this answer by f to find 1 df µ f dt = λ 1 1 = 1 1 T µt T. Separating the derivative gives df f = 1 dt T. Letting the infinitesimal differentials become finite deltas, d, gives us our answer. Incidentally, note that this derivation is general - whenever you have some function y x α, where α is some number, you can always write dy = α dx, which you can y x easily check by integration. (b) If the difference in frequency, f = 5 beats, and f = 500 Hz, then T = f = T f 5 = 0.0, so a % increase in the tension will increase the frequency by 5 beats 500 per second. 8
9. Chapter 1 - Problem 80. As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.00 g/m and a 1.5 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first ind the midpoint of the wire by folding it in half. You then attach on end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.5 kg weight to the end hanging over the pulley. By vibrating the wire and measuring time with your chronometer, you find that the wire s second harmonic frequency is 100 Hz. Next, with the 1.5 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You ind that the pendulum requires 314 seconds to complete 100 oscillations. Pulling out your trusty calculator, you get to work. What value of g will you report back to headquarters? We don t know the length of the string. However, the length of the vibrating part of the wire is half the total distance, so l = L/, where L is the full length of the string. Now, since the wire is vibrating in the second harmonic, then λ = L is the wavelength on the wire. Since λf = v, and v = T/µ, where T = mg is the tension, we can write λ = L = 1 f mg µ L = mg f µ. Now, when we connect the wire to the ceiling and let it oscillate as a pendulum, it s L oscillation period is T = π, which gives L = ( T g π) g. Setting this equal to our value for L above gives ( ) mg T f µ = g g = 4 π f ( ) 4 π m T µ. Plugging in all the numbers gives g = 8.00 m/s, which is a little bit smaller (about 0%) smaller than Earth s gravity. 9
10. Chapter 1 - Problem 83. A water wave is called a deep-water wave if the water s depth is more than one-quarter of the wavelength. Unlike the waves we ve considered in this chapter, the speed of a deep-water wave depends on its wavelength: gλ v = π. Longer wavelengths travel faster. Let s apply this to standing waves. Consider a diving pool that is 5.0 m deep and 10.0 m wide. Standing water waves can set up across the width of the pool. Because water sloshes up and down at the sides of the pool, the boundary conditions require antinodes at x = 0 and x = L. Thus a standing sound water wave resembles a standing sound wave in an open-open tube. (a) What are the wavelengths of the first three standing-wave modes for water in the pool? Do they satisfy the condition for being deep-water waves? Draw a graph of each. (b) What are the wave speeds for each of these waves? (c) Derive a general expression for the frequencies f m of the possible standing waves. you expression should be in terms of m, g, and L. (d) What are the oscillation periods of the first three standing-wave modes? The standing wave wavelengths are given by λ m = L, where m = 1,, 3,. If L = 10.0 m, then this m gives λ 1 = 0 m, λ = 10 m, and λ 3 = 6.67 m. The graphs for each standing wave is seen to the (a) right. In order to be a deep-water wave, the depth has to be more than 1/4 of the wavelength. The first standing wave just barely fits this condition, while the other two certainly do. (b) The wave speeds are calculated from v m = v = 3.95 m/s, and v 3 = 3. m/s. 10 gλ m π, and give v 1 = 5.59 m/s,
(c) Because λ m f m = v m for each wave, then f m = 1 λ m v m = 1 since λ m = L, then we can write m ng f n = 4πL, λ m where we have redefined m n to avoid confusion with mass. (d) Recall that the period of an oscillation is defined as T = 1/f, and so 4πL T n = ng. Plugging in the values gives T 1 = 3.58 s, T =.53 s, and T 3 =.07 s. gλ m π = g πλ m. Now, 11