Physics 41 Homework Set 2 Chapter 16
Conceptual: Q: 3, 6, 8, 11, 16,
2π
, the smaller the k (the coefficient of x), the larger
k
the wavelength. Thus, the rank according to wavelength is: d, a = b = c, e, f
(iii) frequency: the larger omega (coefficient of t) the larger the frequency: f, e, d=b, a=c
(iv) period is inverseley proportional to frequency so reverse (iii)
(ii) Since λ =
3. (i) d=e, f, c, b, a
(v) v =
ω
Therefore the ratio of the t coefficient to the x coefficient gives the velocity:
k
d, b = e, f, 3, 3
6. Higher tension makes wave speed higher. Greater linear density makes the wave move more
slowly.
8. Yes, among other things it depends on. vm ax = ω A = 2π fA =
2π vA
λ
. Here v is the speed of the
wave.
11. Each element of the rope must support the weight of the rope below it. The tension increases
with height. (It increases linearly, if the rope does not stretch.) Then the wave speed v =
T
μ
increases with height.
16. The speed of a wave on a “massless” string would be infinite!
Problems P: 1, 3, 5, 9, 16, 25, 31, 35, 38, 42, 52, 57
1. At t = 0, a transverse pulse in a wire is described by the function y=
6
x2 + 3 where x and y are in meters. Write the function y(x, t) that describes this pulse if it is traveling in the positive x direction with a speed of 4.50 m/s. P16.1
Replace x by x − vt= x − 4.5t
to get
y=
6
⎡( x − 4.5t) 2 + 3⎤
⎣
⎦
3.Two points A and B on the surface of the Earth are at the same longitude and 60.0° apart in latitude. Suppose that an earthquake at point A creates a P wave that reaches point B by traveling straight through the body of the Earth at a constant speed of 7.80 km/s. The earthquake also radiates a Rayleigh wave, which travels along the surface of the Earth in an analogous way to a surface wave on water, at 4.50 km/s. (a) Which of these two seismic waves arrives at B first? (b) What is the time difference between the arrivals of the two waves at B? Take the radius of the Earth to be 6 370 km. P16.3
(a)
The longitudinal wave travels a shorter distance and is moving faster, so it will arrive
at point B first. The transverse takes more time and carries more energy!
(b)
The wave that travels through the Earth must travel
(
)
a distance of
2R sin 30.0° = 2 6.37 × 106 m sin 30.0° = 6.37 × 106 m
at a speed of
7 800 m/s
Therefore, it takes
6.37 × 106 m
= 817 s
7 800 m s
The wave that travels along the Earth’s surface must travel
a distance of
⎛π
⎞
s= Rθ = R ⎜ rad⎟ = 6.67 × 106 m
⎝3
⎠
at a speed of
4 500 m/s
Therefore, it takes
6.67 × 106
= 1 482 s
4 500
The time difference is
5. (a)
Let u = 10π t− 3π x +
π
4
665 s = 11.1 m in
du
dx
= 10π − 3π
= 0 at a point of constant phase
dt
dt
dx 10
=
= 3.33 m s
dt 3
The velocity is in the positive x-direction .
(b)
π⎞
⎛
y( 0.100,0) = ( 0.350 m ) sin ⎜ −0.300π + ⎟ = −0.054 8 m = −5.48 cm
⎝
4⎠
(c)
k=
(d)
vy =
2π
λ
= 3π : λ = 0.667 m
∂y
π⎞
⎛
= ( 0.350)( 10π ) cos⎜ 10π t− 3π x + ⎟
⎝
∂t
4⎠
ω = 2π f = 10π : f= 5.00 H z
vy,m ax = ( 10π )( 0.350) = 11.0 m s
y = ( 0.020 0 m ) sin ( 2.11x − 3.62t)
P16.9
k = 2.11 rad m λ =
2π
= 2.98 m
k
A = 2.00 cm
in SI units
ω = 3.62 rad s
f=
ω
2π
= 0.576 H z
v = fλ =
ω 2π 3.62
=
= 1.72 m s
2π k 2.11
y (mm)
P16.16 (a)
0.2
0.1
0.0
–0.1
–0.2
t=0
0.2
x (mm)
0.4
FIG. P16.16(a)
(b)
2π
= 18.0 rad m
λ 0.350 m
1
1
T= =
= 0.083 3 s
f 12.0 s
k=
2π
=
ω = 2π f = 2π 12.0 s = 75.4 rad s
v = fλ = (12.0 s) ( 0.350 m ) = 4.20 m s
(c)
y = A sin ( kx + ω t+ φ ) specializes to
−3.00 × 10−2 m = 0.200 m sin ( +φ )
at x = 0 , t= 0 we require
so
y ( x, t) =
y = 0.200 m sin (18.0 x m + 75.4t s+ φ )
φ = −8.63° = −0.151 rad
( 0.200 m ) sin (18.0x m
+ 75.4t s− 0.151 rad )
25. An astronaut on the Moon wishes to measure the local value of the free‐fall acceleration by timing pulses traveling down a wire that has an object of large mass suspended from it. Assume a wire has a mass of 4.00 g and a length of 1.60 m, and that a 3.00‐kg object is suspended from it. A pulse requires 36.1 ms to traverse the length of the wire. Calculate gMoon from these data. (You may ignore the mass of the wire when calculating the tension in it.) P16.25
T = M g is the tension;
v=
T
μ
=
Mg
m
L
=
M gL L
=
m
t
speed.
Then,
M gL L2
= 2
m
t
(
)
1.60 m 4.00 × 10−3 kg
Lm
= 1.64 m s2
and g = 2 =
2
−3
Mt
3.00 kg 3.61× 10 s
(
)
is the wave
31. A 30.0‐m steel wire and a 20.0‐m copper wire, both with 1.00‐mm diameters, are connected end to end and stretched to a tension of 150 N. How long does it take a transverse wave to travel the entire length of the two wires? P16.31
The total time is the sum of the two times.
t=
In each wire
L
μ
=L
v
T
Let A represent the cross-sectional area of one wire. The mass of one
wire can be written both as m = ρV = ρ A L and also as m = μL .
Then we have μ = ρ A =
πρd2
4
2 ⎞1 2
⎛ πρd
Thus, t= L ⎜
⎟
⎝ 4T ⎠
(
)
2 ⎤1 2
(
)
2 ⎤1 2
For copper,
⎡ ( π ) 8 920 1.00 × 10−3
(
)
t= ( 20.0) ⎢
⎢
( 4)( 150)
⎣⎢
For steel,
⎡ ( π ) 7 860 1.00 × 10−3
(
)
t= ( 30.0) ⎢
⎢
( 4)(150)
⎣⎢
The total time is
⎥
⎥
⎦⎥
⎥
⎥
⎦⎥
= 0.137 s
= 0.192 s
0.137 + 0.192 = 0.329 s
35. Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string that has a linear mass density of 4.00 × 10–2 kg/m. If the source can deliver a maximum power of 300 W and the string is under a tension of 100 N, what is the highest frequency at which the source can operate? μ = 4.00 × 10−2 kg m
P16.35 A = 5.00 × 10−2 m
Therefore, v =
P=
1 2 2
μω A v :
2
T
μ
P = 300 W
T = 100 N
= 50.0 m s
ω2 =
ω = 346 rad s
ω
= 55.1 H z
f=
2π
2( 300)
2P
=
2
μ A v 4.00 × 10−2 5.00 × 10−2
(
)(
)
2
( 50.0)
38. The wave function for a wave on a taut string is y(x, t) = (0.350 m)sin(10 π t – 3 π x + π /4) where x is in meters and t in seconds. (a) What is the average rate at which energy is transmitted along the string if the linear mass density is 75.0 g/m? (b) What is the energy contained in each cycle of the wave? *P16.38
π⎞
⎛
Comparing y = 0.35sin ⎜ 10π t− 3π x + ⎟ with y = A sin ( kx − ωt+ φ ) = A sin (ωt− kx − φ + π ) we
⎝
4⎠
have k =
(a)
(b)
3π
λ ω 10π s
, ω = 10π s, A = 0.35 m . Then v = fλ = 2π f
= =
= 3.33 m s .
2π k 3π m
m
The rate of energy transport is
1
1
P = μω 2A 2v = 75 × 10−3 kg m
2
2
(
) (10π s)
The energy per cycle is
1
1
Eλ = P T = μω 2A 2λ = 75 × 10−3 kg m
2
2
(
42. *P16.42 The linear wave equation is
∂ 2y
∂x2
then
=
if
∂y
= − bveb( x− vt)
∂t
and
∂t2
∂ 2y
∂t2
= v2
= b2v2eb( x− vt)
∂ 2y
∂x2
( 0.35 m ) 2 3.33 m
) (10π s)
1 ∂ 2y
v2 ∂t2
∂2y
Therefore,
2
2
( 0.35 m ) 2
y = eb( x− vt)
∂y
= beb( x− vt)
∂x
and
∂ 2y
∂x2
= b2eb( x− vt)
, demonstrating that eb( x− vt) is a solution
P16.52 Assuming the incline to be frictionless and taking the
positive x-direction to be up the incline:
∑ Fx = T − M g sin θ = 0 or the tension in the string is
T = M g sin θ
The speed of transverse waves in the string is then
T
μ
=
M g sin θ
m
L
=
M gL sin θ
m
The time interval for a pulse to travel the string’s length is
Δt=
L
m
=L
=
v
M gL sin θ
mL
M g sin θ
.
2π m
= 3.02 J .
3π
52. Review problem. A block of mass M, supported by a string, rests on an incline making an angle θ with the horizontal (Fig. P16.52). The length of the string is L and its mass is m <<M. Derive an expression for the time interval required for a transverse wave to travel from one end of the string to the other. v=
s = 15.1 W