L = 2 λ 2 = λ (1) In other words, the wavelength of the wave in question equals to the string length,

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Merry Stokes
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1 PHY 309 L. Soltions for Problem set # 6. Textbook problem Q.20 at the end of chapter 5: For any standing wave on a string, the distance between neighboring nodes is λ/2, one half of the wavelength. The wave in qestion has three nodes, one at the left end of the string, one in the middle, and one at the right end. Conseqently, the length L of the string is twice the inter-nodal distance, i.e. L 2 λ 2 λ () In other words, the wavelength of the wave in qestion eqals to the string length, λ(this wave) L. (2) By comparison, the fndamental harmonic of the string has only two nodes at the end of the string bt no nodes in the middle. Conseqently, the distance between neighboring nodes of the fndamental harmonic is the length of the whole string, ths λ(fndamental) 2 L λ(fndamental) 2L. (3) Comparing the last two formlae, we immediately see that λ(this wave) λ(fndamental), (4) 2 the wave in qestion has half the wavelength of the fndamental harmonic. Finally, the wavelength and the freqency of any wave are related to the wave speed as f λ. (5) The standing wave in qestion and the fndamental harmonic obtain on the same string, so the wave speed is the same for both waves. Conseqently, eq. (4) for the wavelengths of

2 the two waves imply that their freqencies are related as f(this wave) λ(this wave) 2 2 λ(fndamental) (6) λ(fndamental) 2 f(fndamental), the wave in qestion has twice the freqency of the fndamental harmonic. Indeed, the wave in qestion is the second harmonic of the string. Textbook problem SP.2 at the end of chapter 5: (a) I presme the string is niform, so the linear density m/l is the same for all parts of the string. Since we know the mass of the whole string, the linear density is the ration of that mass to the length of the whole string, m whole L whole kg.25 m kg/m. (7) The playing part of the string between two points where it s fixed to the gitar s deck has a shorter length L 64 cm < L whole, bt it also has a smaller mass m < m whole. We don t know that mass, bt if the string is niform, then the linear density m/l of the playing part shold be the same as the linear density of the whole string, ( m L m whole L whole kg/m. (8) (b) The speed of transverse wave on a string depends on the string s tension force T and linear density m/l, ths T 720 N kg/m 22, 500 m 2 /s 2 50 m/s. (9) 2

3 (c) The fndamental harmonic of the string is a standing wave with two nodes at points where the string is fixed to the deck and no extra nodes in between. Since the distance between nodes of a standing wave is λ/2 one half of the wavelength the fndamental harmonic has λ 2 L (0) where L 64 cm is the length of the playing part of the gitar string, i.e., the distance between the two points where it s fixed to the gitar deck. Therefore, the wavelength of the fndamental harmonic is λ 2L 2 64 cm.28 m. () (d) The freqency of the fndamental harmonic follows from its wavelength λ (fond in part (c)) and wave speed (fond in part (d)): f λ 50 m/s.28 m 7 cycles/s 7 Hz. (2) (e) Higher harmonics of a string are standing waves that have one or more nodes in the middle of the playing part of the string as well as a node at each end (where the string is fixed to the deck). Altogether, the n th harmonic has n + 3, 4, 5,... nodes so the length L of the playing part of the string is divided into n 2, 3, 4,... intervals between the nodes. Since the length of each sch interval is λ/2, the wavelength λ n of the n th harmonic satisfies n λ n 2 L, (3) hence λ n 2L n. (4) For n this formla gives s the wavelength λ 2L of the fndamental harmonic. The 3

4 higher harmonics have shorter wavelengths λ n λ 2L n. (5) The freqencies f n of the higher harmonics follow from their wavelength and the speed of the wave on the string, f n λ n 2L/n n 2L (6) In particlar, the fndamental harmonic # has freqency f 2L, (7) cf. part (d); it is called the fndamental freqency of the string. The higher harmonics have freqencies that are mltiples of the fndamental freqency, f n n f, n 2, 3, 4,... (8) Specifically, the second harmonic jst above the fndamental has one node in the middle of the string, its wavelength λ 2 L λ /2 is a half of the fndamental harmonic s wavelength, and its freqency f 2 2 f is twice the fndamental freqency. For the string in qestion, λ 2.28 m m and f Hz 334 Hz. (9) 4

5 Non-textbook problem #I: The speed of sond in a flid a gas or a liqid is given by a general formla sond B ρ (20) where ρ m/v is the flid s density and B is its blk modls which relates pressre changes to small relative changes of the volme, δp B δv V. (2) When a gas is compressed or rarefied adiabatically fast enogh to prevent the heat from flowing in or ot of the gas its pressre and volme are related as P V γ constant for some adiabatic coefficient γ that depends on the atomic strctre of the gas s molecles: The monoatomic gases sch as helim or argon have γ 5 3, the diatomic gases sch as N 2 or O 2 have γ 7 5.4, while the polyatomic gases sch as carbon dioxide have γ.3. For any sch gas, the adiabatic modls follows from the absolte pressre of the gas, B γ P. (22) Conseqently, the speed of sond in a gas is sond γp ρ. (23) At the reference altitde on Mars (the eqivalent of sea-level on Earth, except Mars does not have any seas), the atmospheric pressre is abot 600 Pascal (less than % of air pressre on Earth, P Earth 00, 000 Pa). Martian atmosphere consists mostly of carbon dioxide (CO 2 ), so its adiabatic coefficient γ is abot.3. And its density dring very cold 5

6 winter nights reaches abot ρ 8 g/m kg/m 3. (24) The speed of sond in Martian atmosphere follows from eq. (23): Note nits: sond γ P ρ Pa m/s. (25) kg/m3 Pa kg/m 3 J/m3 kg/m 3 J/kg (m/s)2. (26) Ths, when we plg into eq. (23) pressre in Pascals and density in kilograms per cbic meter, the speed of sond comes ot in meters per second. Alternative soltion: Sppose I did not give yo the density of Martian atmosphere bt only its temperatre. In this case, yo cold have obtained the density from the ideal gas law P V nrt (27) where T is the absolte temperatre of the gas (in Kelvins), P is its absolte pressre (in Pascals), R 834 J/K/kmol is the niversal gas constant, and n is the amont of gas in kilomols. Here I se kilomols rather than mols becase the mass of n kilomols of gas with moleclar weight is m n (28) in nits of kilograms rather than grams. Hence, the density of gas in kilograms per cbic meter is ρ m V n n RT/P P RT. (29) Note that this density is proportional to the pressre of the gas, so the ratio P / P ρ P RT RT (30) depends only on the temperatre and moleclar weight of the gas, bt does not depend on 6

7 the pressre itself. Conseqently, the speed of sond in the gas sond γp ρ γrt (3) depends only on the the temperatre and chemical formla of the gas bt does not depend on its pressre. Martian atmosphere is made of mostly carbon dioxide CO 2, so it has moleclar weight and adiabatic coefficient γ.3. When this atmosphere is cooled to 70 Kelvin dring a cold winter night on Mars, the speed of sond in it is sond γrt.3 (834 J/K/kmol) (70 K) 44 kg/kmol 460 J/kg 208 m/s. (32) The formla (3) makes clear why the speed of sond on Mars is lower than on Earth: On Mars, the air has lower temperatre than on Earth bt heavier moleclar weight, T Mars < T Earth bt Mars > Earth, hence ( ) γrt Mars < ( ) γrt Earth (33) and therefore the speed of sond ( sond ) γrt on Mars < ( sond ) γrt. (34) on Earth Non-textbook problem #II: Standing waves in pipes are explained in detail in my notes, so I am not going to repeat the theory here. Instead, let me simply refer yo to eqations (notes.33 38) for pipes with two open ends and eqations (notes.43 50) for pipes with one open end and one closed end. 7

8 (a) A pipe with two open ends contains one half-wavelength λ/2 of the lowest harmonic. Conseqently, given the pipe s length L, the lowest harmonic mst have wavelength 2L 2.5 m 3.0 m. (35) In contrast, a pipe with one open end and one closed end contains only one qarterwavelength λ/4 of its own lowest harmonic. pipe is Hence, the lowest harmonic of the second 4L 4.5 m 6.0 m. (36) (b) The n th standing wave mode in a pipe with two open ends fits n half-wavelength intervals inside the pipe, L n λ 2. (notes.35) Hence, the n th harmonic of an open-open pipe of length L has wavelength n 2L n. (37) For the first pipe in qestion, this gives s λ 2L 3.0 m for the first (fndamental) harmonic (cf. eq. (35)), 2 2L 2.5 m (38) for the second harmonic, 3 2L 3.0 m (39) for the third harmonic, etc., etc. 8

9 In contrast, the n th standing wave mode in a pipe with different ends (one open, one closed) fits 2n qarter-wavelength intervals inside the pipe, ths L (2n ) λ 4. (40) Conseqently, the n th harmonic of an open-closed pipe of length L has wavelength n 4L 2n. (4) For the second pipe in qestion which also has L.5 m this gives s λ 4L 6.0 m for the first (fndamental) harmonic (cf. eq. (36)), 2 4L m (42) for the second harmonic, 3 2L 5.2 m (43) for the third harmonic, etc., etc. (c) Given the speed of sond 340 m/s, the freqency f of any sond wave of known wavelength λ can be fond as f /λ. Ths, the first three harmonics of the first pipe (open-open) have freqencies f open open f open open 2 f open open m/s 3.0 m 340 m/s.5 m 340 m/s.0 m 3 Hz, 227 Hz, 340 Hz. Likewise, the first three harmonics of the second pipe (open-closed) have freqencies (44) f open closed f open closed 2 f open closed m/s 6.0 m 340 m/s 2.0 m 340 m/s.2 m 57 Hz, 70 Hz, 283 Hz. (45) 9

10 Textbook qestion Q.2 at the end of chapter 6: Q) Is the electric field of an electromagnetic wave constant in time? A) Hell, no! In any electromagnetic wave, both the electric and the magnetic fields oscillate with the freqency of the wave, E(t) E 0 sin(2πft + φ 0 ), B(t) B0 sin(2πft + φ 0 ), (46) where the amplitdes E 0 and B 0 and the phase φ 0 depend on space coordinates (x, y, z). For different types of waves the space dependence is different for example, a plane wave rnning in x direction has E(t; x, y, z) B(t; x, y, z) ( E 0 sin 2πft 2πx ), λ ( B 0 sin 2πft 2πx ), λ E 0 B 0, E0 and B 0 x axis. while a standing wave has the same phase everywhere bt (x, y, z) dependent amplitdes E ) (x, y, z) and B 0 (x, y, z). Bt the time dependence of the electric and magnetic fields of any EM wave is always as in eq. (46). (47) Textbook qestion Q.4 at the end of chapter 6: Both the microwaves and the light waves are electromagnetic waves, so in vacm they travel at the same speed c , 458 m/s. (48) However, the light waves and the microwaves have very different freqencies and wavelength. The visible light waves have wavelengths in a rather narrow range from abot 700 nanometers for red light to abot 400 nanometers for violet light, i.e., from.4 million to 2.5 million wavelengths in one meter. The freqencies of these waves which obtain from the wavelengths as f c/λ, see the next problem below range from 430 THz for red light to 740 THz for the violet light, i.e., from 430 trillions to 750 trillions cycles per second. 0

11 By comparison, the microwaves have mch longer wavelengths between millimeter and meter, while their freqencies range from 300 MHz to 300 GHz, i.e., from 300 million to 300 billion cycles per second. Textbook problem SP. at the end of chapter 6: (a) Speed of light waves or any other electromagnetic waves in empty space (vacm) is a niversal constant c , 458 m/s m/s (49) (300,000 km/s or abot 86,000 miles/second). Conseqently, in vacm, the wavelength λ of any EM wave is related to its freqency f as λ f c m/s. (50) On the red end of the visible light spectrm, the longest wavelength average hman eye can see is λ red 750 nm (nano-meters). According to eq. (50), this corresponds to lowest visible freqency f red c λ red m/s m Hz, (5) or 400 TeraHertz. On the violet end of the visible light spectrm, the shortest wavelength average hman eye can see is λ red 380 nm (nano-meters). According to eq. (50), this corresponds to highest visible freqency f violet c λ violet m/s m Hz, (52) or 790 TeraHertz.

12 (b) When EM waves propagate throgh some transparent material air, water, glass, whatever, the interactions between the electric and magnetic field and the atoms slow down the EM wave s speed. Ths, in matter, light moves at a slower speed light in material c n < c (53) where n > is the index of refraction of the material. Since the air is rather dilte, its refraction index is little different from to be precise, n air.0003 so the light waves move throgh air at almost the same speed as throgh the vacm, light in air c n air c. (54) Denser materials have higher refraction coefficients, for example n.33 for water and n.5 for many types of glass. Conseqently, the light moves throgh sch materials notably slower than throgh the vacm, light in,water light in,glass c n water c n glass 3 4 c m/s, 2 3 c m/s. (55) PS: The problem ask yo to calclate the speed of light in glass. I have added a similar calclation for the water to illstrate the general rle, bt yo did not have to do it for this homework. (c) When a wave propagates from one medim to another, its freqency does not change it s always the same nmber of cycles per second. For example, if a motor plls a 200 times each second, then the whole string vibrates at freqency f 200 Hz. The vibrating string pshes the air arond it, which creates a sond wave with the same freqency f 200 Hz. And that sond makes yor ear drm vibrates withe the same freqency f 200 Hz. The same works for the light waves: when light moves from one media into another for example, from vacm to glass its freqency does not change. Bt if the speeds of 2

13 light in the two media are different, then the wavelength of the same light wave wold also be different: λ f λ 2 c/n 2 f. (56) The different wavelength make the light change directions as it crosses from one media to another this is called refraction. To calclate the wavelength of light in some material, all we need is the light s freqency f and the speed of light in that material: the wavelength follow from that as λ /f. Ths, given the reslts of parts (a) and (b) of this problem, we immediately have red light λin glass violet light λin glass light in glass f red light light in glass f violet light m/s Hz m/s Hz 500 nm, 253 nm. (57) PS: Since the speed of light in air is almost the same as in vacm, its wavelength in air is almost the same as in vacm. That s why the problem does not specify whether the very red light has wavelength 750 nm in vacm or in air the difference is negligible. Ditto for 380 nm wavelength of the violet light. For simplicity, I assme those wavelength were measred in vacm, bt in practice it wold make very little difference if they were measred in air. 3

PROBLEMS

PROBLEMS------------------------------------------------ - 7- Thermodynamic Variables and the Eqation of State 1. Compter (a) the nmber of moles and (b) the nmber of molecles in 1.00 cm of an ideal gas