1. PHYS
101
Learning
Object:
Sound
Wave
Speed
&
Phase
of
Medium
Damian
Feldman-‐Kiss
12036133
2. Sound
Wave
Speed
&
Phase
of
Medium
Do
sound
waves
travel
faster
through
water
or
through
air?
Make
a
predicLon.
Now
that
you
have
your
predicLon,
lets
solve
a
simple
physics
problem
and
explore
the
science
behind
the
phenomenon!
3. QuesLon
Knowing
that
your
physics
midterm
is
coming
up,
you
want
to
have
some
fun
with
science
to
keep
your
brain
acLve.
You
decide
to
see
how
fast
sound
travels
in
the
dry
air
outside
the
pool
versus
in
the
water.
You
and
your
pal
shout
at
each
other
from
50
m
apart
outside
the
pool.
You
then
head
inside,
get
changed
and
try
again,
but
this
Lme
under
water.
Calculate
the
Lme
it
takes
for
the
sound
waves
to
travel
to
your
pal
in
each
case.
Assume
temperature
is
20˚C
for
both
scenarios.
Some
helpful
values
follow.
Report
your
answer
with
3
sig.
figs.
to
emphasize
the
difference
in
Lmes.
4. Useful
Info
Approximate
velocity
of
a
sound
wave
in
dry
air:
v
=
331
ms-‐1
+
(0.6
m(s
C)-‐1)
x
T
(T
measured
in
Celsius)
v
=
√(B
/
ρ)
See
below
for
soluLon
Bulk
modulus
(Pa)
Density
(kg/
m3)
Water(l)
20˚C
2.2
x
109
1
000
5. SoluLon
Strategy
• What
are
we
given
and
what
do
we
need
to
solve?
• Visualize
problem
• Manipulate
equaLons
and
solve
6. Step
1:
Given
and
unknown
values
BWater(l)
=
2.2
x
109
Pa
ρWater(l)
=
1
000
kg/m3
vWater(l)
=
?
T
=
20˚C
vAir
=
?
d
=
50
m
tAir
=
?
tWater(l)
=
?
7. Step
2:
Draw
a
Picture
Air:
Water(l):
Hey!!
50
m
50
m
Hey!!
8. Step
3:
EquaLons
Velocity
of
a
sound
wave:
v
=
√(B/ρ)
Approximate
velocity
of
a
sound
wave
in
dry
air:
v
=
331
ms-‐1
+
(0.6
m(s
C)-‐1)
x
T
Time:
v
=
d/t
=>
t
=
d/v
9. Step
4:
Solve
Unknown
Values
vWater(l)
=
√(B/ρ)
tWater(l)
=
d/v
=>
d/(√(B/ρ))
=
50m(√(2.2
x
109
Pa
/
1
000
kg/m3))
=
0.0337
s
vsound
wave
in
dry
air
=
331
ms-‐1
+
(0.6
m(s
C)-‐1)
x
T
tAir
=
d/v
=>
d/(331
ms-‐1
+
(0.6
m(s
C)-‐1)
x
20
C)
=
0.146
s
10. Step
5:
Final
Answer
At
20˚C,
it
takes
approximately
0.0337
s
for
sound
waves
to
travel
50
m
in
liquid
water,
and
approx.
0.146
s
in
dry
air.
In
other
words,
sound
waves
travel
approx.
4
Lmes
faster
in
liquid
water
than
in
air
under
these
condiLons.
11. Learning
Goal
1
As
we
just
observed,
the
propagaLon
speed
of
a
sound
wave
depends
on
the
properLes
of
the
medium.
Let’s
examine
this
in
more
detail.
(Dr.’s
Bates
&
Ropler,
2015,
Lecture
17)
12. Learning
Goal
2
–
Why?
A
sound
wave
is
a
longitudinal
wave—as
the
wave
propagates
through
an
elasLc
medium,
the
medium
alternates
between
regions
of
compression
and
rarefacLon.
The
wave
speed
(v)
equals
the
square
root
of
the
bulk
modulus
(B)
divided
by
the
density
of
the
medium
(p):
v
=
√(B/ρ)
The
bulk
modulus
is
defined
as
“the
raLo
of
the
change
in
pressure
divided
by
the
fracLonal
change
in
volume.”
(Hawkes
et
al.,
2014,
p.
425)
13. Learning
Goal
2
–
Why?
The
bulk
modulus
describes
how
resistant
a
medium
is
to
compression.
Liquids
are
nearly
incompressible,
while
gases
are
compressible.
Liquids
are
denser
than
gases.
Sound
waves
generally
propagate
faster
through
a
liquid
medium
than
a
gaseous
medium
because
of
the
difference
in
the
compressibility
and
density
between
the
two
phases.
But
wait,
what
about
the
concomitant
increase
in
density
from
gases
to
liquids?
Would
this
not
compensate
for
the
difference
in
bulk
modulus
between
the
two
phases?
It
turns
out
that
liquids
are
indeed
so
resistant
to
compression
that
this
is
not
the
case.
Lets
look
at
another
quesLon
to
drive
this
point
home.
(Hawkes
et
al.,
2014,
p.
425-‐426)
14. QuesLon
2
Mercury
is
a
very
dense
liquid
(ρ
=
13
534
kg/m3).
Its
bulk
modulus
is
2.85
x
1010
Pa.
Determine
the
speed
of
sound
in
liquid
mercury.
15. SoluLon
v
=
√(B/ρ)
v
=
√(2.85
x
1010
Pa
/
13
534
kg/m3)
v
=
1
450
m/s
The
speed
of
sound
in
liquid
mercury
is
1
450
m/s.
16. References
Dr.’s
Bates
and
Ropler.
2015.
PHYS
101,
202
Lecture
Slides.
Hawkes
et
al.
2014.
Physics
for
ScienLsts
and
Engineers:
An
interacLve
Approach.
Revised
Custom
Volume
1:
PHYS
101.
Custom
Ed
UBC.
