1. A fisherman’s scale stretches2.8 cm when a3.7kg fish hangs from it.(a)What isthe 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) We 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, the amplitudewill be the distance the fish was pulled down from equilibrium: A = 2.5 cm.The frequency of vibration will be f== = 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=Because 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 is 1.00s,(b)What would bethe period of a 1.00-m-long simple pendulum？Solution:(a) We find the length from2 T=;1.002s=which gives L = 0.248 m(b)We find the period from2 T=; 2 2.01 T s==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 of3.5Hz and with amplitude of4.5cm.(a)What is the effective spring constant for this motion?(b)How much energy is involved in this motion?Solution: (a ) We find the spring constant from2πf = (k /m )1/2;2π(3.5 Hz) = [k /(0.240 kg)]1/2, which gives k = 1.2 ⨯ 102 N/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 sameportion 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 gives 21A 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 decrease as 1/r , so the ratio of amplitudes is 2112100.520A r km A r km=== 8. A violin string vibrates at 294Hz when unfingered.At what frequency will 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 270m s .If the 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 have d = 12λ = 12⨯ (2.06 m) = 1.03 m .10. If two successive harmonics of a vibrating string(振动弦) are 280Hz and 350Hz,what is the frequency of the fundamental?Solution: All harmonics are present in a vibrating string: f n = nf 1 , n = 1, 2, 3, ⋯ Thedifference 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.successivefringes on a screen 5.00m away are 6.5cm apart near the center of the pattern.what is the wavelength and frequency of the light?Solution: For constructive interference, the path difference is a multiple of thewavelength:d sin θ = m λ, 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 (m λ/d ) = mL λ/d .For adjacent fringes, ∆m = 1, so we have ∆y = L λ ∆m /d ;0.065 m = (5.00 m)λ(1)/(0.048 ⨯ 10–3 m),which gives λ = 6.24 ⨯ 10–7 m =0.62 μm . The frequency is f = c /λ = (3.00 ⨯ 108m/s)/(6.24 ⨯ 10–7 m) =4.8 ⨯ 1014 Hz .12. The third-order fringe of 610nm light is observed at an angle of 18when theL Unfingered Fingeredlight falls on two narrow slits.How far apart are the slits?Solution: For constructive interference, the path difference is a multiple of thewavelength:d sin θ = m λ, 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 if illuminatedwith 480-nm light? Assume there is air on both sides of the soap film.Solution: For destructive interference, the optical path difference is 2(21)22nL m λλδ=+=+ (m=0,1,2……) The minimum non-zero thickness ismin 11480()1()17922 1.34nm L m nm n λ==⨯⨯=14.A lens appears greenish yellow (570nm λ=is strongest)when white light reflects from it.what minimum thickness of coating(n=1.28) .Solution:With respect to the incident wave, the wave that reflects from the top surface of the coating has optical pathdifference change of2(21)22nL m λλδ=+=+ m = 1, 2, 3, … .The minimum non-zero thickness occurs for m = 1: min 11570()1()22322 1.28nm L m nm n λ==⨯⨯= 15.How far must the mirror 1M in a Michelson interferometer be moved if 750fringes 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 fringeshifts 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 flat glass.When light ofwavelength 670nm is incident normally,25dark lines are observed(with one at each end).How thick is the foil? t πSolution: There is a phase difference for the reflectedt = 12mλ, 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 isd = 12(24)(670 nm) = 8.04 ⨯ 103 nm = 8.04 μm.17.How far must be the mirror1M in a Michelson interferometer be moved 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 thm dark Newton’s ring,as viewed from directlyabove,is given by r where R is the radius of curvature of the curved glass surface and λis the wavelength of light used.Assume that the thickness of the air gap is much less than R at all points and that r R<<.Solution:At a distance r from the center of the lens, the thickness ofthe air space is y, and the phase difference for the reflectedwaves from the path-length difference and the reflection atthe bottom surface isφ = (2y/λ)2π + π.For the dark rings, we haveφ = (2y/λ)2π + π = (2m + 1)π, m= 0, 1, 2, …; ory = 12mλ, m= 0, 1, 2, … .Because m= 0 corresponds to the dark center, m represents the number of the ring. From the triangle in the diagram, we have r2+ (R–y)2 = R2, or r2=2yR–y2˜ 2yR, when y «R,which becomes dr 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 550nm λ= and thant the 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. The separation of the stars isd = L θ = (10 ly)(9.46 ⨯ 1015 m/ly)(7.46 ⨯ 10–7 rad) = 7.1 ⨯ 1010 m20. Monochromatic light(单色光) falls on a slit that is 33.0010mm -⨯wide.If the anglebetween the first dark fringes on either side of the central maximum is 37.0 (dark fringe to dark fringe),what is the wavelength of the 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）(500nm λ=)?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.Whennormally 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 theseparation of the lines fromd sin θ = m λ;d sin 50° = (1)(460 ⨯ 10–9 m), which gives d = 6.0 ⨯ 10–7 m = 600 nm .。