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Material Type: Assignment; Professor: DiDio; Class: General Physics II; Subject: Physics; University: La Salle University; Term: Spring 2000;
Typology: Assignments
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(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
(^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
and
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).
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.
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.
(d) The wave speed is v = λf = (100 cm) (2.0 Hz) = 2.0×10^2 cm/s.
(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.
u =
du dt =^225 π^ sin (πx^ −^15 πt)^.
Squaring this and adding it to the square of 15πy, we have
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.
129 m s. 0.0300 kg m
v = =
new^ new^ new old (^) old old
v v
τ μ τ μ
v = (30 rad/s)/(2.0 m–1) = 15 m/s.
linear mass density of the string, the tension is
2 4 2
2 2 2 2 1 1
180 m/s 120 N 135N. 170 m/s
v v
τ τ
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
(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.
density of the string. Thus,
3
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
(e) Now the situation depicted in Fig. 16-16(b) applies, so P = 0.
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
form. Thus
(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.
3
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
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
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.
2 3
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
n 2 2
nv n f L L
(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):
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.
8 2 max max 2 max (^3)
7.00 10 N m 3.00 10 m s 7800 kg m
v
which is indeed independent of the diameter of the wire.
Thus, using Eq. 16-26, we obtain the tension:
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 ×
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.