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Halliday/Resnick/Walker 7e
Chapter 16 – Waves 1
- (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one- fourth of a period. The period is T = 4(0.170 s) = 0.680 s.
(b) The frequency is the reciprocal of the period:
1.47 Hz. 0.680 s
f T
(c) A sinusoidal wave travels one wavelength in one period:
1.40 m 2.06 m s. 0.680s
v T
- (a) The angular wave number is
(^2 2) 3.49 m 1. 1.80 m
k −
π π = = = λ
(b) The speed of the wave is
( 1.80 m^ )( 110 rad s^ ) 31.5 m s. 2 2
v f
= λ = = = π π
- Let y 1 = 2.0 mm (corresponding to time t 1 ) and y 2 = –2.0 mm (corresponding to time t 2 ). Then we find
kx + 600t 1 + φ = sin−^1 (2.0/6.0)
and
kx + 600t 2 + φ = sin−^1 (–2.0/6.0).
Subtracting equations gives
600(t 1 – t 2 ) = sin−^1 (2.0/6.0) – sin−^1 (–2.0/6.0).
Thus we find t 1 – t 2 = 0.011 s (or 1.1 ms).
- Using v = fλ, we find the length of one cycle of the wave is
λ = 350/500 = 0.700 m = 700 mm.
From f = 1/T, we find the time for one cycle of oscillation is T = 1/500 = 2.00 × 10 –3^ s = 2.00 ms.
(a) A cycle is equivalent to 2 π radians, so that π/3 rad corresponds to one-sixth of a cycle. The
corresponding length, therefore, is λ/6 = 700/6 = 117 mm.
(b) The interval 1.00 ms is half of T and thus corresponds to half of one cycle, or half of 2π rad.
Thus, the phase difference is (1/2)2 π = π rad.
- (a) The amplitude is ym = 6.0 cm.
(b) We find λ from 2 π/λ = 0.020 π: λ = 1.0×10^2 cm.
(c) Solving 2 πf = ω = 4.0π, we obtain f = 2.0 Hz.
(d) The wave speed is v = λf = (100 cm) (2.0 Hz) = 2.0×10^2 cm/s.
(e) The wave propagates in the –x direction, since the argument of the trig function is kx + ωt
instead of kx – ωt (as in Eq. 16-2).
(f) The maximum transverse speed (found from the time derivative of y) is
1 u (^) max 2 fym 4.0 s 6.0 cm 75cm s. = π = π − =
(g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020π(3.5) + 4.0π(0.26)] = –2.0 cm.
- With length in centimeters and time in seconds, we have
u =
du dt =^225 π^ sin (πx^ −^15 πt)^.
Squaring this and adding it to the square of 15πy, we have
u^2 + (15 πy)^2 = (225 π )^2 [sin^2 ( πx − 15 π t) + cos^2 ( πx − 15 π t)]
so that
u = (225π)^2 - (15πy)^2 = 15π 152 - y^2.
Therefore, where y = 12, u must be ± 135 π. Consequently, the speed there is 424 cm/s = 4. m/s.
500 N
129 m s. 0.0300 kg m
v = =
14. From v = τ μ , we have
new^ new^ new old (^) old old
v v
τ μ τ μ
- (a) The wave speed is given by v = λ/T = ω/k, where λ is the wavelength, T is the period, ω is the angular frequency (2π/T), and k is the angular wave number (2π/λ). The displacement has the form y = ym sin(kx + ωt), so k = 2.0 m–1^ and ω = 30 rad/s. Thus
v = (30 rad/s)/(2.0 m–1) = 15 m/s.
(b) Since the wave speed is given by v = τ μ , where τ is the tension in the string and μ is the
linear mass density of the string, the tension is
( )(^ )
2 4 2
τ = μ v = 1.6 × 10 − kg m 15 m s =0.036 N.
16. We use v = τ /μ ∝ τ to obtain
2 2 2 2 1 1
180 m/s 120 N 135N. 170 m/s
v v
= ^ ^ = ^ =
τ τ
- (a) The amplitude of the wave is ym=0.120 mm.
(b) The wave speed is given by v = τ μ , where τ is the tension in the string and μ is the linear
mass density of the string, so the wavelength is λ = v/f = τ μ /f and the angular wave number is
2 0.50 kg m 2 2 100 Hz 141m. 10 N
k f − π = = π = π = λ
(c) The frequency is f = 100 Hz, so the angular frequency is
ω = 2πf = 2π(100 Hz) = 628 rad/s.
(d) We may write the string displacement in the form y = ym sin(kx + ωt). The plus sign is used since the wave is traveling in the negative x direction. In summary, the wave can be expressed as
y = ( 0.120 mm sin) (^ 141m −^1 ) x + ( 628s −^1 )t.
- (a) We read the amplitude from the graph. It is about 5.0 cm.
(b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm and again with the same slope at about x = 55 cm, so
λ = (55 cm – 15 cm) = 40 cm = 0.40 m.
(c) The wave speed is v = τ / μ ,where τ is the tension in the string and μ is the linear mass
density of the string. Thus,
3
3.6 N
12 m/s. 25 10 kg/m
v = (^) − = ×
(d) The frequency is f = v/λ = (12 m/s)/(0.40 m) = 30 Hz and the period is
T = 1/f = 1/(30 Hz) = 0.033 s.
(e) The maximum string speed is
um = ωym = 2πfym = 2π(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s.
(f) The angular wave number is k = 2π/λ = 2π/(0.40 m) = 16 m–1.
(g) The angular frequency is ω = 2πf = 2π(30 Hz) = 1.9×10^2 rad/s
(h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 × 10 –2^ m. The formula for the displacement gives y(0, 0) = ym sin φ. We wish to select φ so that 5.0 × 10 –2^ sin φ = 4.0 × 10 –2. The solution is either 0.93 rad or 2.21 rad. In the first case the function has a positive slope at x = 0 and matches the graph. In the second case it has negative slope and does not match the graph. We select φ = 0.93 rad.
(i) The string displacement has the form y (x, t) = ym sin(kx + ωt + φ). A plus sign appears in the argument of the trigonometric function because the wave is moving in the negative x direction. Using the results obtained above, the expression for the displacement is
y x t ( , ) = ( 5.0 × 10 −^2 m sin (16 m)^ ^ −^1 ) x + (190s −^1 ) t+0.93 .
- Using Eq. 16–33 for the average power and Eq. 16–26 for the speed of the wave, we solve for f = ω/2π:
(e) Now the situation depicted in Fig. 16-16(b) applies, so P = 0.
- The nth resonant frequency of string A is
A n A A
v n f n l L
while for string B it is
, ,
B n B n A B
v n f n f l L
(a) Thus, we see f1,A = f4,B. That is, the fourth harmonic of B matches the frequency of A’s first harmonic.
(b) Similarly, we find f2,A = f8,B.
(c) No harmonic of B would match (^) 3,
A A A
v f l L
- Possible wavelengths are given by λ = 2L/n, where L is the length of the wire and n is an integer. The corresponding frequencies are given by f = v/λ = nv/2L, where v is the wave speed.
The wave speed is given by v = τ μ = τL / M,where τ is the tension in the wire, μ is the
linear mass density of the wire, and M is the mass of the wire. μ = M/L was used to obtain the last
form. Thus
250 N
(7.91 Hz). 2 2 2 (10.0 m) (0.100 kg) n
n L n n f n L M LM
(a) The lowest frequency is f 1 =7.91 Hz.
(b) The second lowest frequency is f 2 = 2(7.91 Hz) =15.8 Hz.
(c) The third lowest frequency is f 3 = 3(7.91 Hz) =23.7 Hz.
- (a) The wave speed is given by
3
7.00 N
66.1m/s. 2.00 10 kg/1.25m
v = = (^) − = ×
(b) The wavelength of the wave with the lowest resonant frequency f 1 is λ 1 = 2L, where L = 125 cm. Thus,
1 1
66.1 m/s 26.4 Hz. 2(1.25 m)
v f = = = λ
41. (a) The wave speed is given by v= τ μ,where τ is the tension in the string and μ is the
linear mass density of the string. Since the mass density is the mass per unit length, μ = M/L,
where M is the mass of the string and L is its length. Thus
(96.0 N) (8.40 m) 82.0 m/s. 0.120 kg
L
v M
(b) The longest possible wavelength λ for a standing wave is related to the length of the string by L = λ/2, so λ = 2L = 2(8.40 m) = 16.8 m.
(c) The frequency is f = v/λ = (82.0 m/s)/(16.8 m) = 4.88 Hz.
- (a) Eq. 16–26 gives the speed of the wave:
2 3
150 N
144.34 m/s 1.44 10 m/s. 7.20 10 kg/m
v
= = = ≈ ×
×
(b) From the Figure, we find the wavelength of the standing wave to be λ = (2/3)(90.0 cm) = 60. cm.
(c) The frequency is
1.44 10 m/s^2 241Hz. 0.600 m
v f
×
λ
44. Use Eq. 16–66 (for the resonant frequencies) and Eq. 16–26 ( v= τ / μ) to find fn:
n 2 2
nv n f L L
which gives f 3 = (3/2L) τ i μ.
(a) When τf = 4 τi, we get the new frequency
(d) The leading edge of the pulse reaches x = 10 cm at t = (10 – 4.0)/5 = 1.2 s. The particle (say, of the string that carries the pulse) at that location reaches a maximum displacement h = 2 cm at t = (10 – 3.0)/5 = 1.4 s. Finally, the trailing edge of the pulse departs from x = 10 cm at t = (10 – 1.0)/5 = 1.8 s. Thus, we find for h(t) at x = 10 cm (with the horizontal axis, t, in seconds):
67. (a) We take the form of the displacement to be y (x, t) = ym sin(kx – ωt). The speed of a point
on the cord is u (x, t) = ∂y/∂t = – ωym cos(kx – ωt) and its maximum value is um = ωym. The wave
speed, on the other hand, is given by v = λ/T = ω/k. The ratio is
m m m m
u y y ky v k
π = = = λ
(b) The ratio of the speeds depends only on the ratio of the amplitude to the wavelength. Different waves on different cords have the same ratio of speeds if they have the same amplitude and wavelength, regardless of the wave speeds, linear densities of the cords, and the tensions in the cords.
68. Let the cross-sectional area of the wire be A and the density of steel be ρ. The tensile stress is
given by τ/A where τ is the tension in the wire. Also, μ = ρA. Thus,
8 2 max max 2 max (^3)
7.00 10 N m 3.00 10 m s 7800 kg m
A
v
×
= = = = ×
which is indeed independent of the diameter of the wire.
- To oscillate in four loops means n = 4 in Eq. 16-65 (treating both ends of the string as effectively “fixed”). Thus, λ = 2(0.90 m)/4 = 0.45 m. Therefore, the speed of the wave is v = fλ = 27 m/s. The mass-per-unit-length is
μ = m/L = (0.044 kg)/(0.90 m) = 0.049 kg/m.
Thus, using Eq. 16-26, we obtain the tension:
τ = v^2 μ = (27)^2 (0.049) = 36 N.
- (a) For visible light
8 14 min (^9) max
3.0 10 m s 4.3 10 Hz 700 10 m
c f (^) −
×
= = = ×
λ ×
and
8 14 max 9 min
3.0 10 m s 7.5 10 Hz. 400 10 m
c f (^) −
×
= = = ×
λ ×
(b) For radio waves
8 min (^6) max
3.0 10 m s 1.0 m 300 10 Hz
c × λ = = = λ ×
and
8 2 max (^6) min
3.0 10 m s 2.0 10 m. 1.5 10 Hz
c × λ = = = × λ ×
(c) For X rays
8 16 min (^9) max
3.0 10 m s 6.0 10 Hz 5.0 10 m
c f (^) −
×
= = = ×
λ ×
and
8 19 max (^11) min
3.0 10 m s 3.0 10 Hz. 1.0 10 m
c f (^) −
×
= = = ×
λ ×
- We refer to the points where the rope is attached as A and B, respectively. When A and B are not displaced horizontal, the rope is in its initial state (neither stretched (under tension) nor slack).
(c) The values of η for which the rope is slack are given below. In the first case, both
displacements are to the right, but point A is farther to the right than B. In the second case, they are displaced towards each other.
m A m B
m A m B
m A B
η x
η
η
ξ = ξ = > ξ = >
ξ = ξ = ξ > ξ = − <
ξ = ξ = ξ = − <
where in the third case B is displaced leftward toward the undisplaced point A.
(d) The first design works effectively to damp fundamental modes of vibration in the two cables (especially in the shorter one which would have an anti-node at that point), whereas the second one only damps the fundamental mode in the longer cable.
