A bat flies toward a wall, emitting a steady sound of frequency 2.00
kHz. This bat hears its own sound plus the sound reflected by
the wall. How fast should bat fly in order to hear a beat
frequency of 10.0 Hz?
To produce 10 Hz beat, the bat hears 2000 Hz from its
own sound plus 2010 Hz coming from the wall. If v is
the magnitude of bat speed, V is a speed of sound and fw
the frequency the wall receives (and reflects)
Bat is moving source and wall is stationary observer.
V
V −v
=
fW 2000 Hz
Bat is moving observer and wall is stationary source.
V +v
V
=
2010 Hz fW
Solve both equations relative to v
V
344m / s
v=
=
= 0.858m / s
401
401
A 2.00 MHz sound wave travels through a pregnant woman’s
abdomen and is reflected from the fetal heart wall of her unborn
baby. The heart wall is moving toward the sound receiver as the
heart beats. The reflected sound is then mixed with the transmitted
sound, and 85 beats per second are detected. The speed of sound in
body tissue is 1500 m/s. Calculate the speed of the fetal heart wall
at the instant this measurement is made.
If v is the magnitude of the heart wall speed, V is a speed
of sound and fh the frequency the heart receives (and
reflects), f0 is original frequency and f’ is reflected back:
Heart is moving observer and device is stationary source.
V +v
fh =
f0
v
Heart is moving source and device is stationary observer.
V
f =
fh
V −v
V +v
2v
f B = f0 − f ' = f0 − f0
= f0
V −v
V −v
'
Beats:
Solving for v:
⎛
fB ⎞
85 Hz
⎞
⎟⎟ = 1500m / s⎛⎜
v = V ⎜⎜
⎟ = 0.0318m / s
⎝ 2 * 2000000 Hz + 85 Hz ⎠
⎝ 2 fo + f B ⎠
The prism shown in fig. has a refractive index of 1.66, and the
angles A are 25 deg. Two light rays m and n are parallel as
they enter the prism. What is the angle between them after
they emerge.
θb
θa
na sin θ a = nb sin θb
θb = arcsin(na sin θ a / nb ) = arcsin(1.66 ⋅ sin(250 ) = 44.60
Angle relative to the horizontal line is:
θ h = θb − 250 = 19.60
Angle between two beams:
θ = 2θ h = 39.20
Light is incident normally on the short face of a 300-600-900 prism.
See Fig. A drop of liquid is placed on the hypotenuse of the
prism. If the index of the prism is 1.62, find the maximum
index that the liquid may have if the light is to be totally
reflected.
na sin θ a = nb sin θb
Total internal reflection occurs if:
na sin θ a = nb
θ a = 600
nb = 1.62 ⋅ sin(600 ) = 1.40
θa
A luminous object is 4.00 m from the wall. You are to use a concave
mirror to project an image of the object on the wall, with the
image 2.25 times the size of the object.
a) How far should the mirror be from the wall?
b) What should its radius of curvature be
c) Draw the rays diagram.
s ' s + 4.00m
a) m = 2.25 = =
s
s
s = 3.2m
s ' = 7.2m
1 1 2
1
1
b)
+ '= =
+
s s R 3.2m 7.2m
R = 4.43m
3.2 m
c)
4.0 m
f
7.2 m
A concave mirror is to form an image of the filament of a headlight
lamp on a screen 8.00 m from the mirror. The filament is 6.00mm
tall, and the image is to be 36.0 cm tall.
a) How far in front of the vertex of the mirror should the filament be
placed?
b) What should be the radius of curvature of the mirror?
Draw the rays diagram.
y ' 360mm
s'
=
= 60.0 =
a) m =
y 6.0mm
s
s=
1 1 2
1
1
+
b) + ' = =
s s R 0.133m 8.0m
R = 0.261m
0.133 m
c)
f
8.0 m
8.0m
= 0.133m
60
A nuclear bomb containing 8.00 kg of plutonium explodes. The
sum of rest masses of the products of the explosion is less than
original mass by one part of 104.
a) How much energy is released in the explosion.
b) If explosion takes place in 4.00 µs what is the average power
developed by the bomb?
−4
(
)
2
a)
E = mc = 8kg ⋅10 ⋅ 3.0 ⋅10 m / s = 7.2 ⋅1013 J
b)
∆E 7.2 ⋅1013 J
19
Power =
=
=
1
.
8
⋅
10
W
−6
∆t 4.0 ⋅10 s
2
8
In a hypothetical nuclear-fusion reactor, two deuterium nuclei
combine or “fuse” to form one helium nucleus. The mass of
deuterium nucleus, expressed in atomic mass units (u) is
2.0136u; and that of a helium nucleus is 4.0015u.
(1u=1.6605⋅10-27 kg).
a)
b)
a)
How much energy is released when 1.0 kg of deuterium
undergoes fusion
The annual consumption of electrical energy in the United
States in of the order of 1.0⋅1019 J. How much deuterium must
react to produce this much energy?
The fraction of the initial mass that becomes energy is
4.0015u
1−
= 6.382 ⋅10−3
2 ⋅ 2.0136u
Energy release per kilogram is
−3
(
)
2
E = mc = 1kg ⋅ 6.382 ⋅10 ⋅ 3.0 ⋅10 m / s = 5.74 ⋅1014 J
2
b)
1.0 ⋅1019 J
5.74 ⋅1014 J
= 17000kg
kg
8
Electrons go through a single slit 150 nm wide and strike a screen
24.0 cm away. You find that at angles of ±20.00 from the
center of diffraction pattern, no electrons hit the screen but
electrons it at all points close to the center.
a) How fast were these electrons moving when they went
through the slit?
b) What will be the next larger angles at which no electrons hit the
screen?
a)
Single slit diffraction
a ⋅sin θ = mλ
First minimum so m=1
λ = a ⋅ sin θ = 150 ⋅10−9 m ⋅ sin 200 = 5.13 ⋅10−8 m
DeBroglie wavelength
λ = h / mv
h
6.626 ⋅10−34 Js
4
v=
=
=
1
.
42
⋅
10
m/s
−31
−8
mλ 9.11 ⋅10 kg ⋅ 5.13 ⋅10 m
Velocity is much lover than speed of light so nonrelativistic formula is OK.
b) Second minimum m = 2
θ 2 = 43.20
a ⋅ sin θ 2 = 2λ
A beam of photons with wavelength 150 nm and a beam
of electrons having the same energy as photons go
through the same slit. You observe the diffraction
pattern on a distant screen and measure that the photons
produce their first dark band at ±20.50 from the
centerline.
a) How wide in the slit?
b) At what angles relative to the centerline will no
electrons be detected?
a) The first dark band is located at
a=
λ
sin θ
=
a ⋅sin θ = λ
150nm
= 428nm
0
sin 20.5
b) First find lambda for electrons
hc
Eele = E pho =
= 1.324 ⋅10−18 J
λ
h
h
λele =
=
= 4.266 ⋅10−10 m
pele
2mEele
a ⋅sin θ = mλele
m=1, θ=0.0570
m=2, θ=0.1140
m=3, θ=0.1710
…….
A bone fragment found in a cave believed to have been inhabited
by early humans contains 0.21 times as much 14C as an equal
amount of carbon in the atmosphere when the organism
containing the bode died. 14C half-life is 5730 years. Find the
approximate age of the fragment.
N = N 0e
− λt
N
= 0.21
N0
t=−
ln(0.21)
λ
5730 y
= − ln(0.21)
= 13000 y
0.693
A patient is given a 4.2 µCi dose of a radioactive isotope with
a half-life of 2.0h. Assuming the entire dose remains in the
body, how much time must elapse before the activity of the
radioactive isotope is 8.5 counts/minute. (this is about three
times the normal background of 3 counts/minute.)
The activity will decrease by a factor of
4.2 ⋅10 −6 Ci ⋅ 3.70 ⋅1010 Bq / Ci
f =
= 1.097 ⋅106
8.5counts / min⋅1 min/ 60 sec
Elapsed time t
2t / 2.0 h = 1.097 ⋅106
t = 40.1h