**Practice Problems**

1.

B /**Transverse wave particle move perpendicular to wave.**

2.

A /**Longitudinal wave particles move parallel to wave.**

3.

D /**Wave is moving right, \ trough is heading toward dot.**

4.

D /**Any hit will generate a wave, but how you hit the beam determines if it is transverse or longitudinal.**

5.

A /**The left sides will interfere constructively and the right sides will interfere destructively.**

6.

B /**Highest when traveling toward the whistle.**

7.

B /**Highest when traveling away from the whistle.**

8.

A & C /**Same hen the child is momentarily stationary.**

9.

Velocity / Frequency / Wavelength340 m/s / 510 s-1 / 340 m/s/510 s-1

= 0.67 m

337 m/s / 337 m/s/3.5 m

= 96 s-1 / 3.5 m

**(75 m)(0.067 s-1)**

= 5.0 m/s / 0.067 s-1 / 75 m

10.

**fbeats = fB – fA = 100 Hz – 85 Hz = 15 Hz**

11. a.

**v = [FT/(m/L)]½ = [(4 x 103 N)/(0.0125 kg/0.5 m)]½ = 400 m/s**

b.

**l1 = 2L/n = 2(0.5 m)/1 = 1 m**

c.

**v = lf \ f = v/l = 400 m/s/1.0 m = 400 s-1**

12. a.

**f’ = f(vw + vo)/(vw – vs) = 600 s-1(340 + 0)/(340 – 34) = 667 s-1**

b.

**f’ = f(vw + vo)/(vw – vs) = 600 s-1(340 + 34)/(340 – 0) = 660 s-1**

c.

**f’ = f(vw + vo)/(vw – vs) = 600 s-1(340 – 0)/(340 + 34) = 545 s-1**

d.

f’ = f(vw + vo)/(vw – vs) = 600 s-1(340 – 34)/(340 + 0) = 540 s-113.

temperature increases, speed / increases / decreasespower increases, loudness / increases / decreases

distance increases, loudness / increases / decreases

frequency increases, sensitivity / increases / decreases

frequency increases, pitch / increases / decreases

14.

fbeats = fB – fA = vw/lA – vw/lBfbeats = 340 m/s/9.00 m – 340 m/s/9.5 m = 2 s-1

15.

16. a.

v = [Ft/(m/L)]½ = [(4500 N)/(0.010 kg/0.5 m)]½ = 474 m/sb.

l1 = 2L = 2(0.5 m) = 1 mv = lf ® f = v/l = 474 m/s/1.0 m = 474 s-1

c.

v = fl = (600 s-1)(1.0 m) = 600 m/sv = [Ft/(m/L)]½ ® 600 m/s = [Ft/(0.010/0.5)]½\ Ft = 7200 N

17.

F2 = nF1 \ F1 = 400 s-1/2 = 200 s-1F4 = nF1 = 4(200 s-1) = 800 s-1

18.

f’ = f(vw + vo)/(vw – vs) = 600 s-1(340 + 0)/(340 – 17) = 632 s-119.

B / Mirrors transpose left and right.20.

C / As long as the mirror above the half way height between your eyes and feet is uncovered.21.

D / The image is formed where the line from the reflected light is extended to the other side of the mirror.22.

B / The light in glass will bend toward normal (ng > na).23.

B / The light would follow the same path if reversed (n1sinq1 = n2sinq2) since 1 & 2 are interchangeable.24.

A / The index for water is different than air \ the light would not bend in the same way (less diffraction).25.

E / Bend toward normal at left interface and away from normal at right interface.26.

A / Bend away from normal at left interface and toward normal at right interface.27.

C / There is no refraction at either interface.28.

B / The light from the fish bends away from normal into your eye \ you see it farther away than it is.29.

A / Directly at the fish because the laser light will follow the same path as visible light.30. a.

fn = f = c/lf = 3 x 108 m/s/600 x 10-9 m = 5 x 1014 s-1

b.

ln = l/nln = 600 nm/1.50 = 400 nm

c.

vn = c/nvn = 3 x 108 m/s/1.50 = 2 x 108 m/s

31.

n1sinq1 = n2sinq2(1.00)(sin65) = (1.33)(sinq2) \ q2 = 43o

32.

sinqc = nlow/nhighsinqc = 1.00/2.42 \ qC = 24.4o

33.

sinqc = nlow/nhighsin37 = 1.00/nhigh \ nhigh = 1.66

34.

velocity / wavelengthv = c/n = 3 x 108 m/s/1.33

v = 2.26 x 108 m/s / ln = l/n = 450 nm/1.33

ln = 338 nm

35. a.

qi = qr = 70ob.

nRsinqR = nisinqi1.50sinqR = 1.00sin70o \ qR = 38.8o

c.

nRsinqR = nisinqi1.50sin38.8 = 1.00sinqi \ qi = 70o

36.

sinqC = n2/n1 = 1.33/1.50 \ qC = 62o37.

f = ½r = ½(10 cm) = 5 cm38.

Lens / MirrorConcave / diverge / converge

Convex / converge / diverge

39.

F / F /Calculated di / Calculated hi

1/do + 1/di = 1/f

1/3 + 1/di = 1/2 \ di = 6 / hi/ho = -di/do

hi/1 = -6/3 \ hi = -2

F / F

F / F

Calculated di / Calculated hi

1/do + 1/di = 1/f

1/1 + 1/di = 1/2 \ di = -2 / hi/ho = -di/do

hi/1 = -(-2)/1 \ hi = 2

40.

F / FCalculated di / Calculated hi

1/do + 1/di = 1/f

1/2 + 1/di = 1/-2 \ di = -1 / hi/ho = -di/do

hi/2 = -(-1)/2 \ hi = 1

41.

positive / converging / real / upright

negative / diverging / virtual / inverted

42.

greater than 1 / The image is larger than the object.negative / The image is upside down.

43.

/ F / FCalculated di / Calculated hi

1/do + 1/di = 1/f

1/3 + 1/di = 1/2 \ di = 6 / hi/ho = -di/do

hi/1 = -6/3 \ hi = -2

F / F

F / F

Calculated di / Calculated hi

1/do + 1/di = 1/f

1/1 + 1/di = 1/2 \ di = -2 / hi/ho = -di/do

hi/1 = -(-2)/1 \ hi = 2

44.

F / FCalculated di / Calculated hi

1/do + 1/di = 1/f

1/2 + 1/di = 1/-2 \ di = -1 / hi/ho = -di/do

hi/2 = -(-1)/2 \ hi = 1

45. a.

1/do + 1/di = 1/f1/20 + 1/di = 1/15 \ di = 60 cm

b.

hi/ho = -di/dohi = -hodi/do = (-1.5 cm)(60/20) = -4.5 cm

46.

make-up mirror / Concave (converging)passenger side mirror / Convex (diverging)

47.

do / real/virtual / inverted/upright / |M|do > 2f / real / inverted / < 1

do = 2f / real / inverted / = 1

2f > do > f / real / inverted / > 1

do = f / no / no / no

0 < do < f / virtual / upright / > 1

48. a.

b.

image distance, di / magnification, M / image height, hi1/do + 1/di = 1/f

1/20 + 1/di = 1/10

di = 20 cm / M = -di/do

M = -20/20 = -1 / M = hi/ho

-1 = hi/2.0

hi = -2.0 cm

c.

real / virtual / inverted / upright / larger / same size / smaller49. a.

F / Fb.

image distance, di / magnification, M / image height, hi1/do + 1/di = 1/f

1/10 + 1/di = 1/-10

di = -5 cm / M = -di/do

M = -(-5)/10 = ½ / M = hi/ho

½ = hi/2.0

hi = 1.0 cm

c.

real / virtual / inverted / upright / larger / same size / smaller50. a.

/ F / Fb.

image distance, di / magnification, M / image height, hi1/do + 1/di = 1/f

1/30 + 1/di = 1/20

di = 60 cm / M = -di/do

M = -60/30 = -2 / M = hi/ho

-2 = hi/1.0

hi = -2.0 cm

c.

real / virtual / inverted / upright / larger / same size / smaller51. a.

Which side is converging in function? / rightWhich side could be used as a make-up mirror? / right

Which side could be used as a side mirror in a car? / left

Which point is closest to focus? / 3

b.

an upright magnified image? / 2an inverted unmagnified image? / 5

an inverted image that is smaller than the object? / 6

no image at all / 3

an upright image that is smaller than the object? / 1

an inverted image that is larger than the object? / 4

52. a.

b.

Virtual, it forms on the same side as the object.c.

1/do + 1/di = 1/f1/6 + 1/di = 1/10 \ di = -15 cm

d.

hi/ho = -di/do = -(-15)/6 = 2.5e.

1/do + 1/di = 1/f1/20 + 1/di = 1/10 \ di = 20 cm

(M = -di/do = -20/20 = -1)

The image is on the left side at 20 cm. The image is inverted and the same size as the object.

53. a.

Ab.

B and Ec.

C and D54. a. __tan__q = x/L.

b. __sin__q = l/d.

c.

qo / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10sin / .017 / .035 / .052 / .070 / .087 / .105 / .122 / .139 / .156 / .174

tan / .017 / .035 / .052 / .070 / .087 / .105 / .123 / .141 / .158 / .176

d. 6o

e.

tanq = x/L and sinq = ml/d / x = mlL/dtanq = x/L and sinq = (m + ½)l/d / x = (m + ½)lL/d

55.

Order D

_1__ 1½l

_1__ __1l

_0__ _½l

_0__ _0l

_0__ _½l

_1__ __1l

_1__ 1½l

56.

The intensity decrease as m increases.57. a.

sinq = ml/dsinq = 2(6 x 10-7 m)/(0.50 x 10-3 m) = 2.4 x 10-3\q = 0.14o

b.

sinq = (m+ ½)l/dsinq = ½(6 x 10-7 m)/(0.50 x 10-3 m) = 6.0 x 10-4 \q = .034o

c. (1)

tanq = x/Lx = Ltanq = (2.0 m)tan0.034 = 1.2 x 10-3 m

(2)

x = (m + ½)lL/dx = ½(600 x 10-9 m)(2.0 m)/(0.500 x 10-3 m) = 1.2 x 10-3 m

d.

W = 2x = 2(1.2 x 10-3 m) = 2.4 x 10-3 me. (1)

W » 2lL/d » 2(600 x 10-9 m)(2.0 m)/(0.0005 m) = .0048 m(2)

W » 2lL/d » 2(600 x 10-9 m)(2.0 m)/(0.001 m) = 0.0024 m(3)

The spot light is wider than it is long.58. a.

tanq = x/Ltanq = 0.284 m/1.00 m \ q = 15.9o

b.

d = Length(m)/linesd = 0.001 m/500 grooves = 2 x 10-6 m

c.

sinq = ml/dl = dsinq/m = (2 x 10-6 m)sin15.9o/1 = 5.46 x 10-7 m

d.

ml/d = x/Ll = dx/mL = (2 x 10-6)(0.284)/(1)(1.00) = 5.68 x 10-7 m

e.

Part c is correct because the q 6o59. a.

ln = l/n = 500 nm/1.25 = 400 nmb.

nfilm is middle and reflection is dark \T = ¼l = ¼(400 nm) = 100 nm

c.

No, under water, nfilm is extreme \ ¼l would produce constructive interference.60. a.

ln = l/n = 450 nm/1.35 = 333 nmb.

nfilm is extreme and reflection is bright \T = ¼l = ¼(333 nm) = 83 nm

61. a.

tanq = x/L = 0.12 x 10-2 m/1 m = 1.2 x 10-3\ q = 0.069o

b.

x/L = ml/d ® l = xd/mLl = (0.12 x 10-2 m)(0.5 x 10-3 m)/(1)(1 m) = 600 nm

c.

x/L = (m + ½)l/dx = (1 m)(0.5)(600 x 10-9 m)/(0.5 x 10-3 m) = 6 x 10-4 m

d.

Width = 2xW = 2(6 x 10-4 m) = 1.2 x 10-3 m

e.

x is proportional to wavelength \ increase62. a.

tanq = (30.5 x 10-2 m)/(1.00 m) \ q = 17.0osinq = ml/d = l/d

l = dsinq = (1 x 10-3/600)sin17.0o = 4.86 x 10-7 m

b.

tanq = (42.8 x 10-2 m)/(1.00 m) \ q = 23.2osinq = ml/d = l/d

l = dsinq = (1 x 10-3/600)sin23.2o = 6.56 x 10-7 m

63.

Bright reflection and lfilm is extreme \ T = ¼lfilmT = ¼(l/n) = 750 nm/(4)(1.35) = 140 nm

64.

dark reflection and lfilm is middle \ T = ¼lfilmT = lfilm/4 = l/4n = 550 nm/(4)(1.22) = 113 nm

Practice Multiple Choice

1.

B / A full wavelength is between two corresponding points \ a half wavelength is half as long.2.

D / The entire diagram includes two complete waves, which is 6 m long \ 1 wavelength = 3 m.3.

D / v = lf = (3 m)(9 s-1) = 27 m/s4.

C / The velocity depends on tension and mass/length of the rope, vw = [FT/(m/L)]½, not on amplitude.5.

B / Look for 2.0 Hz on the x-axis, go straight up to the curve and to the left to determine the wavelength.6.

B / v = lf = (2.5 m)(2.0 s-1) = 5 m/s7.

A / Air can only support longitudinal waves because air particles are not attached to each other.8.

B / Amplitudes are added when waves meet, which is minimized where crest meets trough.9.

A / In a transverse wave, the particle move at right angles to the wave motion.10.

B / This is the second harmonic (number of loops).11.

D / v = lf = (2 m)(5 s-1) = 10 m/s12.

B / fn = nf1 ® 5 s-1 = (2)f1 \ f1 = 2.5 s-113.

A / fn = nf1 ® f = 3f1 \ f1 = f/314.