双语物理

1. A fisherman’s scale stretches

2.8 cm when a

3.7kg fish hangs from it.(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) 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 amplitude

will be the distance the fish was pulled down from equilibrium: A = 2.5 cm.

The frequency of vibration will be f

=

= = 3.0 Hz

2. 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 have

f2/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 be

the period of a 1.00-m-long simple pendulum？

Solution:(a) We find the length from

2 T=

;1.002

s=which gives L = 0.248 m

(b)We find the period from

2 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 is

ω = 2πf = 2π(264 Hz) = 1.66 ? 103 s–1.

(a) The maximum speed is

v max = ωA = (1.66 ? 103 s–1)(1.5 ? 10–3 m) =2.5 m/s.

(b) The maximum acceleration is

a 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 ) 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 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 have

22211()I A I A =; 221

2()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 is

I 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 the

final 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 Hz

9. 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, ? 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 is the wavelength and frequency of the light?

Solution: For constructive interference, the path difference is a multiple of the

wavelength:

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 ? 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

L Unfingered Fingered

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 = m λ, m = 0, 1, 2, 3, … .

For the third order, we have

d 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 soap

film(n=1.34) that would appear black if illuminated

with 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)22

nL m λλδ=+=+ (m=0,1,2……) The minimum non-zero thickness is

min 11480()1()17922 1.34

nm 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 path

difference change of

2(21)22nL m λλδ=+=+ m = 1, 2, 3, … .

The minimum non-zero thickness occurs for m = 1: min 11570()1()22322 1.28

nm L m nm n λ==??= 15.How far must the mirror 1M in a Michelson interferometer be moved if 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 is

m ? = ?d 2 /λ;

344 = 2(0.125 ? 10–3 m)/λ, which gives λ = 7.27 ? 10–7 m = 727 nm

16. A fine metal foil separates one end of two pieces of optically flat glass.When light of

wavelength 670nm is incident normally,25dark lines are observed(with one at each end).How thick is the foil? t π

Solution: There is a phase difference for the reflected

t = 1

2

mλ, m= 0, 1, 2, … .

Because m = 0 corresponds to the edge where the glasses

touch, m+ 1 represents the number of the fringe.

Thus the thickness of the foil is

d = 1

2

(24)(670 nm) = 8.04 ? 103 nm = 8.04 μm.

17.How far must be the mirror

1

M 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 is

N = 2 ?L/λ;

750 = 2 ?L/(589 ? 10–9 m), which gives ?L = 2.21 ? 10–4 m = 0.221mm

18. Show that the radius r of the th

m dark Newton’s ring,as viewed from directly

above,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 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 is

φ = (2y/λ)2π + π.

For the dark rings, we have

φ = (2y/λ)2π + π = (2m + 1)π, m= 0, 1, 2, …; or

y = 1

2

mλ, 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 d

r 2 = 2(12

m 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 is

θ = 1.22λ/D = (1.22)(550 ? 10–9 m)/(0.90 m) = 7.46 ? 10–7 rad. The separation of the stars is

d = 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 is 33.0010mm -?wide.If the angle

between 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 is

θ = 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 .