oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
Question 1, chap 9, sect 99.
part 1 of 1
10 points
Given: G = 6.6726 × 10−11 N m2 /kg2
Three 2 kg masses are located at points in
the x-y plane as shown in the figure.
1
and has magnitude
mm
f2 = G 2
y
G m2
=
y2
(6.6726 × 10−11 N m2 /kg2 )(2 kg)2
=
(0.59 m)2
= 7.66745 × 10−10 N .
59 cm
f2
43 cm
What is the magnitude of the resultant
force (caused by the other two masses) on
the mass at the origin?
Correct answer: 1.6345 × 10−9 N (tolerance
± 1 %).
Explanation:
Let : m = 2 kg
x = 43 cm ,
y = 59 cm .
and
Basic Concepts: Newton’s Law of Gravitation:
m1 m2
Fg = G
r2
We calculate the forces one by one, and
then add them using the superposition principle. The force from the mass on the right
is pointing in the x direction, and has magnitude
mm
x2
G m2
=
x2
(6.6726 × 10−11 N m2 /kg2 )(2 kg)2
=
(0.43 m)2
= 1.4435 × 10−9 N .
f1 = G
The other force is pointing in the y direction
F
θ
f1
Now we simply add the two forces, using
vector addition. Since they are at right angles
to each other, however, we can use Pythagoras’ theorem as well:
q
F = f12 + f22
h
= (1.4435 × 10−9 N)2
i 21
+ (7.66745 × 10−10 N)2
= 1.6345 × 10−9 N .
As you see, this force is very small. If the
masses in the picture are standing on a table
(the view being from above), typically the
force of static friction will not be overcome.
In agreement with common sense, then, the
masses will not move.
Note: The angle θ shown in the figure is
f2
θ = arctan
f1
(7.66745 × 10−10 N)
= arctan
(1.4435 × 10−9 N)
= 27.9759 ◦ .
Question 2, chap 9, sect 1.
part 1 of 1
10 points
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
Planet X has four times the diameter and
three times the mass of the earth.
What is the ratio of gravitational acceleration at the surface of planet X to the gravitational acceleration at the surface of the Earth,
gx
?
ge
9
gx
=
1.
ge
64
gx
3
2.
=
ge
8
gx
3
3.
=
correct
ge
16
gx
1
4.
=
ge
4
gx
4
5.
=
ge
9
4
gx
=
6.
ge
25
gx
9
7.
=
ge
49
gx
7
8.
=
ge
16
gx
5
9.
=
ge
49
1
gx
=
10.
ge
5
Explanation:
Let : MX = 3 Me ,
RX = 4 Re .
Because
G Me m
m ge =
,
Re2
where Me is the mass of the earth and Re
G Me
. Similar
is the radius. So that ge =
Re2
equation applies to the planet X
gx =
G Mx
.
Rx2
The ratio between the surface gravitational
accelerations is
Mx Re2
gx
=
ge
Me Rx2
=
3 Me Re2
Me (4 Re )2
=
3
.
16
2
Question 3, chap 9, sect 99.
part 1 of 1
10 points
An astronaut weighs 185 N on the Moon’s
surface. He is in a circular orbit about
the Moon at an altitude of seven-eighth the
moon’s radius.
What gravitational force does the Moon
exert on him?
Correct answer: 52.6222 N (tolerance ± 1
%).
Explanation:
7
Rmoon ,
8
7
Rastronaut = 1 +
Rmoon ,
8
W = m g = 185 N .
Let : h =
The astronaut’s weight on the Moon is
w = m gmoon = 185 N ,
where the gravitational acceleration gmoon on
the Moon’s surface is
gmoon = G
Mmoon
.
2
Rmoon
At a height h = Rmoon above the surface, the
free-fall acceleration is
gmoon = G gmoon
2 .
2 = 15
225
Rmoon
8
64
Mmoon
Therefore, the gravitational force the Moon
exerts on the orbiting astronaut is
64
W
225
64
(185 N)
=
225
= 52.6222 N .
F =
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
Question 4, chap 18, sect 4.
part 1 of 1
10 points
A heavy liquid with a density 11 g/cm3 is
poured into a U-tube as shown in the lefthand figure below. The left-hand arm of the
tube has a cross-sectional area of 15.5 cm2 ,
and the right-hand arm has a cross-sectional
area of 4.98 cm2 . A quantity of 96.9 g of a
light liquid with a density 0.92 g/cm3 is then
poured into the right-hand arm as shown in
the right-hand figure below.
15.5 cm2 4.98 cm2 15.5 cm2 4.98 cm2
3
After the light liquid has been added to the
right side of the tube, a volume A2 h2 of heavy
liquid is displaced to the left side, raising the
heavy liquid on the left side by a height of
h1 with a displaced volume of A1 h1 . Since
the volume of heavy liquid is not changed, we
have
A1 h1 = A2 h2 .
At the level of the heavy-light liquid interface
in the right side, the absolute pressure is
P = Patm + ρℓ g L ,
and at the same level in the left tube,
P = Patm + ρh g [h1 + h2 ] .
Equating these two values, we obtain
L
h1
ρℓ g L = ρh g [h1 + h2 ]
A1
= ρh g h1 +
h1
A2
A1
.
= ρh g h1 1 +
A2
h2
Solving for h1 , we have
ρ L
ℓ
A1
ρh 1 +
A2
(0.92 g/cm3 ) (21.1498 cm)
=
2) (15.5
cm
(11 g/cm3 ) 1 +
(4.98 cm2 )
h1 =
heavy liquid
11 g/cm3
light liquid
0.92 g/cm3
If the density of the heavy liquid is
11 g/cm3 , by what height h1 does the heavy
liquid rise in the left arm?
Correct answer: 0.430131 cm (tolerance ± 1
%).
= 0.430131 cm .
Explanation:
Let :
mℓ
A1
A2
ρℓ
ρh
= 96.9 g ,
= 15.5 cm2 ,
= 4.98 cm2 ,
= 0.92 g/cm3 ,
= 11 g/cm3 .
Question 5, chap 18, sect 5.
part 1 of 1
10 points
and
Using the definition of density
mℓ
mℓ
=
ρℓ =
Vℓ
A2 L
mℓ
L=
= 21.1498 cm .
A2 ρ ℓ
Note: Mercury is denser than steel, and
both are denser than water. Mercury and
water do not mix.
Consider a steel ball floating on the surface
of mercury in a half-filled container.
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
What happens when the rest of the container
is filled with water?
1. The ball sinks to the bottom of the container
2. The ball remains partially submerged in
mercury to exactly the same depth as before.
4
correct
Explanation:
Mercury is much denser than the water, so
it stays in the bottom half of the container
while water fills the top half. The steel ball
floats in mercury but sinks in water, so it
has to float at the mercury-water interface,
partially submerged in mercury and partially
in water.
The only non-trivial question is whether
adding water makes the ball go deeper into
mercury or less deep. To answer this question,
note that the ball floating at the interface is
subject to two buoyant forces:
in
BM = ρM g Vmercury
3. The ball remains partially submerged in
mercury to a greater depth then before.
and
in
BW = ρW g Vwater
,
hence the equilibrium condition is
Mball g = BM + BW
in
= ρM g Vmercury
+
4. Mercury floats on top of water, and the
ball floats on mercury’s surface
Although the water is much less dense then
mercury, it does provide a bit of buoyant force,
hence in equilibrium
in
Vmercury
[after] <
5. The ball floats to the surface of the water.
(1)
in
ρW g Vwater
.
Mball
ρM
(2)
after adding the water.
On the other hand, before adding the water,
the whole weight of the ball was balanced by
the mercury’s buoyant force, hence
Mball g = BM + 0
6. The ball remains partially submerged in
mercury but not as deeply as before.
and therefore
in
Vmercury
[before] =
before adding the water.
Mball
ρM
(3)
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
Comparing eqs. (2) and (3), we immediately see that
in
in
Vmercury
[after] < Vmercury
[before],
(4)
which means that after adding the water, the
ball floats less deep in mercury than before.
The ball remains partially submerged in
mercury but not as deeply as before.
5
To a good approximation, water is incompressible (at ordinary pressures) and thus
must obey the continuity equation
2
d
F = Av = π
v = constant.
2
In other words,
d22 v2 = d21 v1 ,
and hence the speed of the water flow through
the constricted segment of the pipe is
v2 = v1
Question 6, chap 18, sect 6.
part 1 of 1
10 points
Given: Atmospheric pressure is 101300 Pa.
The density of water is 1000 kg/m3 . The
viscosity of water is negligible.
Water flows at speed of 4.2 m/s through a
horizontal pipe of diameter 3.4 cm. The gauge
pressure P1 of the water in the pipe is 1.6 atm.
A short segment of the pipe is constricted to
a smaller diameter of 2.3 cm.
1.6 atm
P2
P1
d21
d22
(3.4 cm)2
(2.3 cm)2
= 9.17807 m/s .
= (4.2 m/s)
In the absence of viscosity, the water pressure and speed obey the Bernoulli equation
P + ρ g h + 21 ρ v 2 = constant,
which for a horizontal pipe (h1 = h2 ) gives us
P2 + 21 ρ v22 = P1 + 21 ρ v12 ,
or
4.2 m/s
3.4 cm
v2
2.3 cm
What is the gauge pressure of the water
flowing through the constricted segment?
Correct answer: 1.27129 atm (tolerance ± 1
%).
Explanation:
Let : Patm = 101300 Pa ,
ρ = 1000 kg/m3 ,
η = 0,
v1 = 4.2 m/s ,
d1 = 3.4 cm ,
d2 = 2.3 cm ,
gauge
P2
= 1.6 atm ,
P2 = P1 − 21 ρ v22 − v12 .
Strictly speaking, the pressures P1 and P2 in
the Bernoulli equation are the absolute pressures rather than gauge pressures. However,
since the air pressure outside the pipe is constant, the difference P2gauge − P1gauge is the
same as the difference between the absolute
pressures P2 − P1 . Therefore,
1
ρ v22 − v12
2
1000 kg/m3
= (162080 Pa) −
2
2
× (9.17807 m/s) − (4.2 m/s)2
= [(162080 Pa) − (33298.5 Pa)]
× (9.87167 × 10−6 atm/Pa)
P2gauge = P1gauge −
= 1.27129 atm .
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
6
s
37 L
8. T = 2 π
Question 7, chap 11, sect 3.
42 g
part 1 of 1
10 points
s
29 L
9. T = 2 π
Consider a light rod of negligible mass and
35 g
s
length L pivoted on a frictionless horizontal
41 L
bearing at a point O . Attached to the end of
10. T = 2 π
correct
the rod is a mass M1 . Also, a second mass M2
45 g
of equal size(i.e., M1 = M2 = M ) is attached
Explanation:
1
L from the lower end , as
to the rod
Basic Concepts: The momentum of iner5
tia is
shown in the figure below.
I ≡ M d2 .
In this case there are two masses with
O
L
1
5 L
M2
θ
M1
The period of this pendulum in the small
angle approximation is given by
s
13 L
1. T = 2 π
15 g
s
17 L
2. T = 2 π
20 g
s
65 L
3. T = 2 π
72 g
s
89 L
4. T = 2 π
104 g
s
53 L
5. T = 2 π
63 g
s
85 L
6. T = 2 π
99 g
s
25 L
7. T = 2 π
28 g
IM1 = M L2
2
1
IM2 = M L − L
5
2
4
L
=M
5
16
M L2 , we get
=
25
I = IM1 + IM2
16
= 1+
M L2
25
41
=
M L2 .
25
(1)
Torque: “τ ≡ r F sin φ” The relationship
between torque and angular acceleration is
“τ = I α′′ .
We have F = (M1 + M2 ) g exerted on the
center of mass, and the distance between O
and the center of mass is
r1 M + r2 M
2M
4
LM + LM
5
=
2M
9
L.
=
10
rcm ≡
(2)
And φ = −θ. Then (for both masses), using
the small angle approximation, we get
τ = −2 M g rcm sin θ ≃ −2 M g rcm θ.
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
Using Eq. 2, the torque is given by
τ = −2 M g rcm θ
9
= −2 M g
L θ
10
9
= − M gLθ.
5
(3)
From this expression we find the Newton’s
second law, gives us
9
τ = I α = − M gLθ,
5
where α =
d2 θ
dt2
7
What is the maximum amplitude of the
motion for which the block does not separate
from the plate?
Correct answer: 0.1984 m (tolerance ± 1 %).
Explanation:
Basic Concepts: Newton’s 2nd law:
X
F = ma
Simple harmonic motion:
x = A cos(ω t)
(4)
d2 x
= −A ω 2 cos(ω t) .
dt2
Solution: At each instant, there are two
forces acting on the block: the force of gravity
Fg = m g and the normal force N from the
plate. The block separates from the plate
when N = 0. Newton 2nd law reads
a=
, so
d2 θ
9 M gL
=
−
θ.
dt2
5
I
The expression for simple harmonic motion
α=
d2 θ
= −ω 2 θ ,
dt2
−m g + N = m a ,
(5)
where up is positive. We see that
and Eqs. 2 through 4, when compared to Eq.
5, yields
2 M g rcm
,
ω =
I
2π
T ≡
ω s
2
N = m (g + a)
so for N to vanish, we must have
so
I
2 M g rcm
v u 41
u
M L2
u
u 25
= 2 πu t 9
M gL
5
s
41 L
= 2π
.
45 g
= 2π
amax = −g ,
(6)
Question 8, chap 15, sect 1.
part 1 of 1
10 points
A block rests on a flat plate that executes
vertical simple harmonic motion with a period
of 0.894 s.
The acceleration of gravity is 9.8 m/s2 .
(1)
(the normal force N also vanishes for any
acceleration quicker than this, but we are
only looking for the point when the block
just barely leaves the plate). We also know
that for simple harmonic motion
a = −A ω 2 cos(ω t) ,
and at the highest and lowest points, the acceleration is at a maximum. The acceleration
of the block is positive (up) at the lowest
point, and negative (down) at the highest
point. We want the negative one, so
amax = −Amax ω 2 .
Comparing with equation (1), we note
Amax ω 2 = g ,
or
Amax =
g
,
ω2
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
and using T =
2π
, we find
ω
g T2
4 π2
(9.8 m/s2 ) (0.894 s)2
=
4 π2
= 0.1984 m .
Amax =
Question 9, chap 15, sect 99.
part 1 of 1
10 points
The motion of a piston in an auto engine
is simple harmonic. The piston travels back
and forth between the extreme points 23 cm
apart.
Question 10, chap 16, sect 2.
part 1 of 1
10 points
At t = 0, a transverse wave pulse in a wire
is described by the function
y = 5 e−x
1. y = 5 e−x
2
/2
2. y = 5 e−x
2
/2
− 4.5 t
3. y = 5 e−x
2
/2
+ 4.5 t
5. y = 5 e−(x
23 cm
What is the maximum speed of the piston when the engine is running at 10870 rpm
(revolution per minute)?
Correct answer: 130.905 m/s (tolerance ± 1
%).
Explanation:
The maximum speed is
v = ωA,
where A is the oscillation amplitude and ω is
the angular frequency. In this case, A is half
the distance between the two extreme points.
So
A = 11.5 cm = 0.115 m .
The angular frequency of SHM is
rev 2 π rad
1 min
ω = 10870
min
1 rev
60 s
= 1138.3 rad/s .
Combining yields
v = ωA
= (1138.3 rad/s) (0.115 m)
= 130.905 m/s .
2
/2
,
where x and t are given in SI units.
Which of the following functions describes
this wave if it is traveling in the negative x
direction with a speed of 4.5 m/s .
4. y = 5 e−(x+4.5 t)
10870 rpm
8
2
2
/2
correct
−4.5 t)/2
6. y = 5 e−(x−4.5 t)
7. y = 5 e−(x
2
2
/2
+4.5 t)/2
Explanation:
For the wave traveling in the negative direction with the speed of 4.5 m/s , we need to
replace x with x + 4.5 t which gives
y = 5 e−(x+4.5 t)
2
/2
.
Question 11, chap 16, sect 4.
part 1 of 1
10 points
Tension is maintained in a 31 m string as in
the figure. The observed wave speed is 21 m/s
when the suspended mass is 3.5 kg .
The acceleration of gravity is 9.8 m/s2 .
31 m
21 m/s
3.5 kg
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
What is the wave speed when the suspended
mass is 1.4 kg ?
Correct answer: 13.2816 m/s (tolerance ± 1
%).
Explanation:
Let : ℓ = 31 m ,
v1 = 21 m/s ,
m1 = 3.5 kg , and
m2 = 1.4 kg .
When m1 = 3.5 kg, the tension is
T1 = m1 g = 34.3 N, and
s
T1
v2 =
.
µ
When m2 = 1.4 kg, the tension is
T2 = m2 g = 13.72 N, and
s
T2
v2 =
µ
r
T2
v1
=
T1
r
m2
=
v1
m2
s
(1.4 kg)
(21 m/s)
=
(3.5 kg)
= 13.2816 m/s .
Question 12, chap 17, sect 2.
part 1 of 1
10 points
A rifle is fired in a valley with parallel vertical walls. The echo from one wall is heard
1.72 s after the rifle was fired. The echo from
the other wall is heard 3.61 s after the first
echo.
The velocity of sound is v = 343 m/s .
How wide is the valley?
Correct answer: 1209.08 m (tolerance ± 1
%).
Explanation:
9
Let : t1 = 1.72 s ,
t2 = 3.61 s , and
v = 343 m/s .
Let the distance from the closer wall be d1
and the distance from the farther wall be d2 .
The first echo is heard after traveling a
distance of 2 d1 = v t1 . The second echo
is heard after traveling a distance of 2 d2 =
v (t1 + t2 ). Each time the sound traveled from
your position to the wall and back to you.
2 d = v t1 + v (t1 + t2 ) = v (2 t1 + t2 )
Thus the distance
dibetween these walls is
h
d=
=
v 2 t1 + t2
2
h
i
(343 m/s) 2 (1.72 s) + (3.61 s)
2
= 1209.08 m .
Question 13, chap 17, sect 2.
part 1 of 1
10 points
The sound level β in decibels is defined to
be
I
β ≡ 10 log ,
I0
−12
where I0 = 1 × 10
W/m2 .
The decibel scale intensity for busy traffic is
79 dB. Two people having a loud conversation
have a decibel intensity of 70 dB.
What is the approximate combined sound
intensity?
Correct answer: 8.94328 × 10−5 W/m2 (tolerance ± 1 %).
Explanation:
Inverting the definition of sound level in
decibels gives
I = I0 10β/10 .
Therefore, the combined sound intensity I is
I = I1 + I2
i
h
= I0 10β1 /10 + 10β2 /10
h
i
= (1 × 10−12 W/m2 ) 107.9 + 107
= 8.94328 × 10−5 W/m2 .
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
Question 14, chap 17, sect 4.
part 1 of 1
10 points
A helicopter drops a paratrooper carrying
a siren that emits a 883 Hz audible signal.
The microphone (reciever) on the plane monitors the signal from the transmitter as the
paratrooper falls.
Take the speed of sound in air to be 343 m/s
and assume the paratrooper always remains
below the helicopter.
If the perceived frequency becomes constant at 424 Hz, what is the terminal speed of
the paratrooper?
Correct answer: 371.314 m/s (tolerance ± 1
%).
Explanation:
Let :
f = 883 Hz ,
f1 = 424 Hz , and
va = 343 m/s .
Because the paratrooper is moving away from
the helicopter,
va
f1 = f
, so
va + vs
f
vs =
− 1 va
f1
883 Hz
− 1 (343 m/s)
=
424 Hz
= 371.314 m/s .
Question 15, chap 16, sect 4.
part 1 of 1
10 points
The distance between two nearest nodes of
a standing wave is 23.5 cm . Hand generated
pulses move up and down through a complete
cycle seven times every eight seconds.
Find the velocity of the wave.
Correct answer: 0.41125 m/s (tolerance ± 1
%).
Explanation:
10
The distance between two nodes is half the
wavelength, so
λ = 2d.
The frequence is the number of cycles per unit
time, so
n
f= .
t
The velocity is given by
v=fλ
2nd
=
t
2 (7) (23.5 cm)
=
(8 s)
= 0.41125 m/s .
Dimensional analysis for v:
cm
1m
cycles · cm
=
·
= m/s
s
s 100 cm
Question 16, chap 16, sect 4.
part 1 of 2
10 points
The length of a string is 564 cm . The tension in the string is 28 N . The linear density of
the string is 0.0095 kg/m . The string is held
fixed at each end. The string vibrates in three
sections; i.e., the string has three antinodes.
Determine the frequency of vibrations in
the string.
Correct answer: 14.4387 Hz (tolerance ± 1
%).
Explanation:
Let : N = 3 = number of antinodes
v = 54.2897 m/s , and
ℓ = 564 cm .
When a string is held fixed at each end, the
wavelength is
2ℓ
, where N = 1, 2, 3, · · ·
λ=
N
1m
2 (564 cm)
·
=
3
100 cm
= 3.76 m .
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
The velocity of wave on the string is
s
T
v=
µ
s
28 N
=
0.0095 kg/m
= 54.2897 m/s ,
so the frequency of the wave is
v
λ
54.2897 m/s
=
3.76 m
= 14.4387 Hz .
f=
Determine the wavelength of the sound
wave in this hollow pipe.
2ℓ
5
2ℓ
2. λ =
7
1. λ =
3. λ = ℓ
4. λ =
5. λ =
6. λ =
7. λ =
Question 17, chap 16, sect 4.
part 2 of 2
10 points
What is the fundamental frequency; i.e.,
the lowest frequency the string can sustain?
Correct answer: 4.81291 Hz (tolerance ± 1
%).
8. λ =
Explanation:
Basic Concepts: When both ends of a
pipe are open, the wavelength is
λ=
v
λf
v
=
2ℓ
54.2897 m/s
=
2 (5.64 m)
= 4.81291 Hz .
Question 18, chap 17, sect 3.
part 1 of 2
10 points
The figure below represents a sound wave
in a hollow pipe with both ends open.
2ℓ
9
2ℓ
3
ℓ
correct
4
ℓ
3
ℓ
2
9. λ = 2 ℓ
Explanation:
ff =
2ℓ
,
N
where
N = 1, 2, 3, 4, · · · ,
(1)
where N is the number of nodes.
When one end of a pipe is closed and the
other end open, the wavelength is
λ=
4ℓ
, where
2N − 1
N = 1, 2, 3, · · · , (2)
since there is node at one end.
Solution: There are eight nodes (N = 8)
in the air column.
If the number of quarter wavelengths is
even, both ends have either nodes or antinodes; however, if the number of quarter
wavelengths is odd, one end has a node and
the other end has an anti-node.
λ
The number J of quarter wavelengths in
4
the length of the pipe ℓ is J = 16 , since
J =
ℓ
11
ℓ
λ
4
or
λ=
4ℓ
4ℓ
=
.
J
16
oldmidterm 04 – JYOTHINDRAN, VISHNU – Due: Apr 28 2007, 4:00 am
The number of quarter wavelengths is even.
We will use Eq. 1 to solve for the number of
nodes.
2ℓ
,
N
2ℓ
=
8
ℓ
=
.
4
λ=
where
N =8
Question 19, chap 17, sect 3.
part 2 of 2
10 points
Consider another organ pipe which has one
end open and one end closed.
wavelengths is odd, one end has a node and
the other end has an anti-node.
λ
The number J of quarter wavelengths in
4
the length of the pipe ℓ is J = 15 , since
J =
1. λ =
2. λ =
3. λ =
4. λ =
5. λ =
6. λ =
7. λ =
4ℓ
3
4ℓ
11
4ℓ
13
4ℓ
9
4ℓ
5
4ℓ
correct
15
4ℓ
7
8. λ = 4 ℓ
4ℓ
17
Explanation:
Solution: There are eight nodes (N = 8)
in the air column.
If the number of quarter wavelengths is
even, both ends have either nodes or antinodes; however, if the number of quarter
9. λ =
ℓ
λ
4
or
λ=
4ℓ
4ℓ
=
.
J
15
The number of quarter wavelengths is odd.
We will use Eq. 2 to solve for the number of
nodes.
4ℓ
,
2N −1
4ℓ
=
2 (8) − 1
λ=
=
ℓ
Determine the wave length of the sound
wave in this hollow pipe.
12
4ℓ
.
15
where
N =8