2012-2013第一学期复习题[1]

1.A fisherman’s scale stretches2.8 cm when a3.7kg fish hangs fromit.(a)What is the spring constant?(b)What will be the amplitude and frequency of vibration if the fish is pulled down 2.5cm more and released so that it vibrates up and down?Solution: (a) W e find the spring constant from the elongation caused by the weight:k=mg/x∆=(3.7 kg)(9.80 m/s2)/(0.028 m)= 1.30⨯103 N/m.(b) Because the fish will oscillate about the equilibrium position, theamplitude will be the distance the fish was pulled down from equilibrium: A = 2.5 cm.The frequency of vibration will be f =2kmπ=31.30103.72Nmkgπ⨯= 3.0 Hz2. A mass m at the end of a spring vibrates with a frequency of 0.88Hz;when an additional 1.25kg mass is added to m,the frequency is 0.48Hz.What is the value of m?The dependence of the frequency on the mass is f =2k m πBecause the spring constant does not change, we havef2/f = (m/m2)1/2;(0.48 Hz)/(0.88 Hz) = [m/(m + 1.25 kg)]1/2, which gives m = 0.53 kg.3. (a)Determine the length of a simple pendulum whose period is1.00s,(b)What would be the period of a 1.00-m-long simple pendulum？Solution:(a) We find the length from2L T gπ=;21.0029.8L s msπ= which gives L = 0.248 m(b )We find the period from2L T gπ=; 21.0022.019.8m T s msπ==4. A tuning fork vibrates at a frequency of 264Hz and the tip of each prong moves 1.5mm to either side of center.Calculate(a)the maximum speed and (b)the maximum acceleration of the tip of a prong. Solution: The angular frequency of the motion isω = 2πf = 2π(264 Hz) = 1.66 ⨯ 103 s –1.(a ) The maximum speed isv max = ωA = (1.66 ⨯ 103 s –1)(1.5 ⨯ 10–3 m) =2.5 m/s. (b ) The maximum acceleration isa max = ω2A = (1.66 ⨯ 103 s –1)2(1.5 ⨯ 10–3 m) =4.1 ⨯ 103 m/s 2.5.A mass of 240g oscillates on a horizontal frictionless surface at a frequency of 3.5Hz and with amplitude of 4.5cm.(a)What is the effective spring constant for this motion?(b)How much energy is involved in this motion?Solution: (a ) W e find the spring constant from2πf = (k /m )1/2;2π(3.5 Hz) = [k /(0.240 kg)]1/2, which gives k = 1.2 ⨯ 102N/m .(b ) We find the total energy from the maximum potential energy:E = U max =212kA=12(1.16 ⨯ 102 N/m)(0.045 m)2 = 0.12 J .6. Two earthquake waves have the same frequency as they travel through the same portion of the Earth,but one is carrying twice the energy.What is the ratio of the amplitudes of the two waves?Solution: Because the speed, frequency, and medium are the same for the two waves, the intensity depends on the amplitude only: 2I A∝For the ratio of intensities we have22211()I A I A =;2212()A A =, which gives21A A = 1.41.7.Compare(a)the intensities and (b)the amplitudes of an earthquake P wave as it passes two points 10km and 20km from the source.Solution:We assume that the wave spreads out uniformly in all directions. (a ) The intensity will decrease as 1/r 2, so the ratio of intensities isI 2/I 1 = (r 1/r 2)2 = [(10 km)/(20 km)]2 =0.25.(b ) Because the intensity depends on 2A, the amplitude will decreaseas 1/r , so the ratio of amplitudes is2112100.520A r km A r km===8. A violin string vibrates at 294Hz when unfingered.At what frequencywill it vibrate if it is fingered one-fourth of the way down from the end? Solution:From the diagram the initial wavelength is 2L , and thefinal wavelength is 3L /2. The tension has not changed,so the velocity has not changed:v = f 1λ1 = f 2λ2;(294 Hz)(2L ) = f 2(3L /2), which gives f 2 = 392 Hz9. The velocity of waves on a string is 270ms.Ifthe frequency of standing waves is 131Hz,how far apart are the nodes? Solution: We find the wavelength from v = f λ; 270 m/s = (131 Hz)λ, which gives λ = 2.06 m.The distance between adjacent nodes is 12λ, so we haved =12λ =12⨯ (2.06 m) = 1.03 m .10. If two successive harmonics of a vibrating string(振动弦) are 280Hz and350Hz,what is the frequency of the fundamental?Solution: All harmonics are present in a vibrating string: f n = nf 1 , n = 1, 2, 3, ⋯ The difference in frequencies for two successive harmonics is ∆f = f n +1 – f n = (n + 1)f 1 – nf 1 = f 1 , so we have f 1 = 350 Hz – 280 Hz = 70 Hz . Note that the given harmonics correspond to n = 4 and 5.11.monochromatic light falls on two very narrow slits 0.048mm apart.successive fringes on a screen 5.00m away are 6.5cm apart near the center of the pattern.what isthe wavelength and frequency of theLUnfingeredFingeredlight?Solution: For constructive interference, the path difference is a multiple of the wavelength:d sin θ = mλ, m= 0, 1, 2, 3, … .We find the location on the screen from y = L tan θ.For small angles, sin θ≈ tan θ, which gives y = L(mλ/d) = mLλ/d.For adjacent fringes, ∆m = 1, so we have ∆y = Lλ ∆m/d;0.065 m = (5.00 m)λ(1)/(0.048 ⨯ 10–3 m),which gives λ = 6.24 ⨯ 10–7m =0.62 μm. The frequency is f= c/λ=(3.00 ⨯ 108 m/s)/(6.24 ⨯ 10–7 m) =4.8 ⨯ 1014 Hz.12. The third-order fringe of 610nm light is observed at an angle of 18 when the light falls on two narrow slits.How far apart are the slits? Solution: For constructive interference, the path difference is a multiple of the wavelength:d sin θ = mλ, m= 0, 1, 2, 3, … .For the third order, we haved sin 18° = (3)(610 ⨯ 10–9 m), which gives d = 5.9 ⨯ 10–6 m =5.9 μm.13.what is the minimum thickness(>0)of a soapfilm(n=1.34) that would appear black ifilluminated with 480-nm light? Assume thereis air on both sides of the soap film.Solution: For destructive interference, the optical path difference is2(21)22nL m λλδ=+=+ (m=0,1,2……)The minimum non-zero thickness ism in 11480()1()179221.34nm L m nm nλ==⨯⨯=14.A lens appears greenish yellow (570nm λ=is strongest )when whitelight reflects from it.what minimum thickness of coating(n=1.28) . Solution:With respect to the incident wave, the wave that reflectsfrom the top surface of the coating has optical path difference change of 2(21)22nL m λλδ=+=+ m = 1, 2, 3, … .The minimum non-zero thickness occurs for m = 1:m in 11570()1()223221.28nm L m nmnλ==⨯⨯=15.How far must the mirror1Min a Michelson interferometer be movedif 750 fringes of 589-nm light are to pass by a reference line? One fringe shift corresponds to a change in path length of λ. The number of fringe shifts produced by a mirror movement of ∆L ism∆ = ∆d 2 /λ;344 = 2(0.125 ⨯ 10–3 m)/λ, which gives λ = 7.27 ⨯ 10–7 m = 727 nm16. A fine metal foil separates one end of two pieces of optically flattφ1 =πφ2 =(2t /λfilm )2π+πglass.When light of wavelength 670nm is incident normally,25dark lines are observed(with one at each end).How thick is the foil? Solution: There is a phase difference for the reflected t = 12m λ, m = 0, 1, 2, … .Because m = 0 corresponds to the edge where the glassestouch, m + 1 represents the number of the fringe. Thus the thickness of the foil is d =12(24)(670 nm) = 8.04 ⨯ 103 nm = 8.04 μm .17.How far must be the mirror1Min a Michelson interferometer bemoved if 750 fringes of 589-nm light are to pass by a reference line? Solution: One fringe shift corresponds to a change in path length of λ. The number of fringe shifts produced by a mirror movement of ∆L isN = 2 ∆L /λ;750 = 2 ∆L /(589 ⨯ 10–9 m), which gives ∆L = 2.21 ⨯ 10–4 m = 0.221mm18. Show that the radius r of the thmdark Newton ’s ring,as viewed from directly above,is given byr m Rλ=where R is the radius of curvatureof the curved glass surface andλis the wavelength of light used.Assumethat the thickness of the air gap is much less than R at all points and thattdLr R<<.Solution:At a distance r from the center of the lens, the thickness of the air space is y , and the phase difference for the reflected waves from the path-length difference and the reflection at the bottom surface isφ = (2y /λ)2π + π.For the dark rings, we haveφ = (2y /λ)2π + π = (2m + 1)π, m = 0,1, 2, …; or y =12m λ, m = 0, 1, 2, … .Because m = 0 corresponds to the dark center, m represents the number of the ring. From thetriangle in the diagram, we have r 2 + (R – y )2 = R 2, or r 2 = 2yR – y 2 ˜ 2yR , when y « R ,which becomesr 2 = 2(12m λ)R = m λR , m = 0, 1, 2, … ; or r = (m λR )1/2.19. Two stars 10 light-years away are barely resolved by a 90cm(mirror diameter)telescope.How far apart the stars?Assume 550nmλ= and thantthe resolution is limited by diffraction. Solution: The resolution of the telescope isθ = 1.22λ/D = (1.22)(550 ⨯ 10–9 m)/(0.90 m) = 7.46 ⨯ 10–7 rad.RyrThe separation of the stars isd = L θ = (10 ly)(9.46 ⨯ 1015 m/ly)(7.46 ⨯ 10–7 rad) = 7.1 ⨯ 1010 m 20. Monochromatic light(单色光) falls on a slit that is33.0010mm-⨯wide.Ifthe angle between the first dark fringes on either side of the central maximum is37.0(dark fringe to dark fringe),what is the wavelength ofthe light used?Solution: The angle from the central maximum to the first minimum is 18.5°.We find the wavelength from a sin θ1min = m λ;(3.00 ⨯ 10–6 m) sin (18.5°) = (1)λ, which gives λ = 9.52 ⨯ 10–7 m = 952 nm.21. What is the angular resolution limit set by diffraction(衍射) for the 100-inch(mirror diameter)Mt.Wilson telescope （威尔逊望远镜）(500nmλ=)?Solution: The minimum angular resolution isθ = 1.22λ/D = (1.22)(500 ⨯ 10–9 m)/(100 in)(0.0254 m/in) =2.4 ⨯ 10–7 rad = (1.4 ⨯ 10–5)°22. The wings of a certain beetle have a series of parallel lines across them.When normally incident 460-nm light is reflected from the wing,the wing appears bright when viewed at an angle of 50 .How far apart are the lines?Solution: The lines act like a grating. Assuming the first order, we find the separation of the lines from d sin θ = m λ;d sin 50° = (1)(460 ⨯ 10–9 m), which gives d = 6.0 ⨯ 10–7 m = 600 nm .。