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Ch 14 – Waves and Sound
14.1 Types of Waves
All mechanical waves require
 source of disturbance and
 a disturbable medium that can influence
itself
Point element (the black point) below shows an
object in the water where a wave passes by.
The object rises up then goes back down as the
wave passes.
There are two type of waves
Transverse Waves: A traveling wave or pulse that
causes the elements of the disturbed medium to
moved perpendicular to the direction of
propagation.
As you can see below…the distance traveled by
a wave from a point is Δx which equals vt.
We can say the distance traveled by a wave in a
given about of time is, vt.
So the position of any wave is given by the
starting position (x) and the distance traveled
after time (vt).
Longitudinal Waves: A traveling wave or pulse that
causes the elements of the medium to move parallel Travels to the right:
to the direction of the propagation.
Travels to the left:
y(x,t) = f(x – vt)
y(x,t) = f(x + vt)
Earthquakes are an example of both types of waves.
Primary (P) waves are the longitude waves which travel faster than the
transverse or Secondary (S) waves.
The longitudinal waves travel about 7.5 km/s near the surface and the transverse waves travel near
the surface through the earth at 4.5 km/s.
Example:
Ocean waves with a crest-to-crest distance of 10.0 m
and v = 1.20 m/s can be described by the wave function
y(x, t) = (0.800 m) sin[0.628(x – vt)]
(a) Sketch y(x, t) at t = 0.
(b) Sketch y(x, t) at t = 2.00 s
Note how the entire wave form has shifted 2.40 m in
the +x direction in this time interval.
Note: sine or cosine can be used for harmonic wave
motion, this example uses sine, most in your book use
cosine.
Longnitudnal & Transverse Waves: OW-B-LT
14.2 Waves on a String
v2 = Ftension / μ;
F=m a
F = m v2/r
The radial (centripetal) force is
Tension(sinθ)
Fr = FT sinθ
(by one side)
The rope is being pulled on two sides so
total radial force is given by
Fr = 2 FT sinθ
If θ is small (<10°) then sinθ = θ
Fr = 2 FT θ
2 FT θ
2 FT θ
FT
v2
= (m) v2/r
= (2 μ R θ) v2/r
= ( μ ) v2
= FT / μ
Mass associated with Δs is μΔs
(where μ is mass per unit length)
m = μ (Δs) Δs = 2(R θ)
m = μ (2 R θ)
m = 2 μ R θ
Example
A piano string having a mass per unit length 10.0  10–3 kg/m is under a
tension of 1000 N. Find the speed of a wave traveling on this string.
14.2 continues
v2
v2
v
= FT / μ
= 1000 / 0.010
= 316 m/s
When a rope fixed to
an immoveable object
reflects off the object,
the wave reflects off in
the opposite direction,
with the negative
amplitude.
If the wave continues to
another medium, some of
the energy will be
transmitted to the new
different sized medium,
and some of the energy will
be reflected.
Newton’s 3rd Law…for every action there is an
opposite and equal reaction.
The total energy will
remain the same.
(13.5: T = ½kA2)
Why in the case of the high density rope to low
density rope is the reflected wave not inverted?
How would the above
picture be different
if instead of being
attached to the wall,
it was attached to a
freely moving ring
Similar explanation as the rings…the low density (thin)
rope resists change much less than the high density
rope. Thus the low density rope accelerates up to the
new Amplitude, then the downward tension pulls the
rope back down.
(free to move up and
down with the same
amplitude as the wave)?
Ans:
For the thin rope to the thick rope…the thick rope
resists change similarly to the rope fixed to the wall.
Waves on a Rope: OW-B-WR
Wave Machine with Rods: OW-B-WM
14.3 Harmonic Wave Functions
A sinusoidal wave is created by plotting sin θ versus θ.
Crest
(+ max amplitude)
In the form of
y(x, t) = A cos (a x)
Trough
(- max amplitude)
If the wave is moving to the right then
y(x, t) = A cos [a (x – vt)]
Wavelength
(length of complete wave, until
start of repetition)
frequency = 1/T
measured in hertz
v=λ/T
k = 2π / λ ; ω = 2π / T
v = ω/k
and the wave function is generally expressed as
y(x, t) = A cos [kx – ωt + φ]
Amplitude = A
v = -ωAsin(kx – ωt);
vmax = ωA
Wave # is k
a = -ω2Acos(kx – ωt);
amax = ω2A
Example
A sinusoidal wave on a string is described by
y(x, t) = A cos [ k x – ω t + φ]
y(x, t) = 0.51 cos [310 x – 9.30 t]
(3.1 rad/cm = 310 rad/m)
y = (0.51 cm) sin(kx – t)
where k = 3.10 rad/cm and  = 9.30 rad/s.
How far does a wave crest move in 10.0 s?
Does it move in the (+) or (-) x direction?
14.4 Sound Waves
v= ω /k
v = 9.3/310
v = 0.030 m/s
vave = ∆x / ∆t
0.030 = ∆x / 10.0
∆x = 0.300 meters (+x-dir)
Type of sound wave:
 Audible
 Infrasonic
 Ultrasonic
The speed of sound is given below. The first is the speed of sound in air. The
latter is the speed of sound in liquids and solids.
331 1 
Tc
v
273
Find vsound in mercury, with a
bulk modulus of about 2.80 
1010 N/m2 and a density of
13600 kg/m3.
Find vsound in an Aluminum
rod, which has a Young’s
modulus of about 7.0  1010
N/m2 and a density of 2700
kg/m3.
elastic _ prop

inertial _ prop
v
v
B


stress / strain


Y

2.80  1010
 1.43 km s
13.6  103
7.0  1010 Pa
m

 5090

s
2.70  103 kg 3
m
Y
Flaming Standing Waves: OW-D-SF
14.5 Sound Intensity
The piston transmits energy to the element of air in the tube. This energy is propagated away from
the piston by the sound wave. We must use the velocity to determine the total kinetic energy in one
wavelength.
v(x,t) = -ω smaxsin(kx - ωt)
How would we define intensity?
Suppose you take a large magnifying glass, go outside on a nice sunny day
ΔK = Δ½ mv2
ΔK = ½ Δm (ω smax)2sin2(kx - ωt)
If t is constant (at a given instant in time)
ΔK = ½ Δρ V
(ω smax)2sin2(kx)
dK
= ½Δρ A dx (ω smax)2sin2(kx)
K
K
K
Kλ
= ½ΔρA (ω smax) sin (kx)dx
= ½ΔρA (ω smax)2
½ x (see note below)
= ½ΔρA (ω smax)2
½λ
2
= ¼ ρA(ωsmax) λ
2
2
The total potential energy for one wavelength is
the same as the kinetic
so the total mechanical energy is
E λ = K λ +U λ = ½ ρA(ωsmax)2λ
And the rate of energy transfer is the power of
the wave which is the energy that passes by a
given point during one period of oscillation
P = Work/time
P = Eλ / T
P = ½ ρA(ωsmax)2 (λ / T)
P = ½ ρA(ωsmax)2 (v)
P = ½ ρAv(ωsmax)2
NOTE explanation
Double angle formulas
cos(2x) = 1 – 2sin2(x) solve for sin
sin2(x) = ½ - ½ cos(2x)
near June 20th and play with ants. Would agree the intensity of the light
rays that strike the ants is intense?
The intensity I of a wave is defined as the power per unit
area,
I = Power / A = Power / 4πr2
This is the rate at which the energy being transported by the wave
transfers through a spherical shell unit of radius, r, and area, A,
perpendicular to the direction of the wave. The intensity decreases in
proportion to the square of the distance from the source.
I=
Power
/A
I = ½ ρAv(ωsmax)2 / A
I = ½ ρv(ωsmax)2
In terms of pressure amplitude
(ΔPmax = ρvωsmax)
P
=
F / A
P
=
m
a
/ A
P
= ρ(A*Δx) Δv/Δt / A
P
= ρ (Δx) Δv/Δt
P
=ρ
(Δx)/Δt
Δv
Pmax = ρ -ω smaxsin(kx - ωt) Δv
ΔPmax = ρ ω smax
(1)
Δv
ΔPmax = ρvωsmax
I = ΔP2 / 2ρv
A logarithmic scale is best used to determine the intensity
level,
β = 10 log (I / Io)
I0, 1.00 x 10-12 W/ m2, is the reference intensity also the
threshold of hearing at 1000 Hz
Threshold of pain
Threshold of hearing
Now we can integrate
∫sin2(x) dx = ∫½dx - ½∫cos(2x)dx
∫sin2(x) dx = ½x – ¼ ∫sin(2x) + const
If x >> 1 then simplifies to ½ x
Example
What is the sound level that
corresponds to an intensity
of 10 x 10-7 W/m2 ?
I = 1.00 W/m2
β = 120 dB
I0 = 1.00 x 10-12 W/ m2
β = 0 dB
β = 10 log (
I
/
Io
)
β = 10 log (10 x 10-7 W/m2 / 1.0 x 10-12 W/m2)
β = 10 log 106
Rule of thumb: A doubling in the loudness is
β = 60 dB
approximately equivalent to an increase of 3 dB
Range of Hearing: OW-C-RH
Example
A loudspeaker is placed between
two observers who are 30 m apart,
along the line connecting them. If
one observer records a sound level
of 60.0 dB, and the other records a
sound level of 66.0 dB, how far is
the speaker from each observer?
Derive the equation
β2 – β1 = 20 log (r1 / r2) from
β = 10 log (I / I0)
β2 – β1 = 20 log (r1 / r2)
66 – 60 = 20 log (r1 / r2)
0.3 = log (r1 / r2)
2 r2 = r1
10 and 20 meters
β1 = 10 log (I1 / I0);
β2 = 10 log (I2 / I0)
14.6 The Doppler Effect
The Doppler effect is the apparent change in
frequency (or wavelength) that occurs (because of
motion of the source or observer) of a wave.
When the relative speed of the source and
observer is higher than the speed of the wave,
the frequency appears to increase (if lower, then
freq appears to decrease)
We also call an appearance of increased frequency, a blue shift
or blue-shifted…and decreased frequency, a red shift. Why?
 low energy, freq
high energy, high freq 
Radio micro IR R OYGB IV UV X
γ
β2 β2 β2 β2 -
β1 = 10
β1 = 10
β1 = 10
β1 = 10
log(I2/I0) - 10 log(I1/I0)
[log (I2/I0) - log(I1/I0)]
log [ (I2 / I0) / (I1 / I0)]
log (
I2
/ I1 )
P2 = P1
β2 - β1 = 10 log (P2/4πr22 / P1/4πr12)
β2 - β1 = 10 log ( r1 / r2) 2
β2 - β1 = 20 log ( r1 / r2)
A convenient graphical representation is to use
circular arcs concentric to the source which the
fronts of which are called wave fronts
When the observer moves toward the source, the
speed of the waves relative to the observer is
Actual wavelength doesn’t change so
f’ / v’ = f / v
f’ = f (v’/ v)
Toward:
Away:
v ’ = v + vo
vo is the observer speed
The frequency heard by the observer appears
higher when the observer approaches the source
f’ = f ( v’ ) / v
v ’ = v - vo
vo is the observer speed
The frequency heard by the observer appears lower
when the observer moves away from the source
f’ = f ( v’ ) / v
f’ = f (v + vo)/ v
f’ = f (v - vo)/ v
Example
At the Winter Olympics, an athlete rides her luge
down the track while a bell just above the wall of the
chute rings continuously. When her sled passes the
bell, she hears the frequency of the bell fall by the
musical interval called a minor third. That is, the
frequency she hears drops to five sixths of its
original value. (a) Find the speed of sound in air at
the ambient temperature, 10.0C. (b) Find the speed
(a)
v = 331 + 0.6*TC
v = 331 + 0.6(-10);
(b) f’’ / f’ = 5/6
Approaching bell
f’ = f (v + vo)/ v
5 / 6 = f (v - vo)/ v
5 / 6 = (v - vo)
v = 325 m/s
Leaving bell
f’’ = f (v - vo)/ v
/ f (v + vo)/ v
/ (v + vo)
of the athlete.
5(v + vo) = 6 (v - vo)
vo = 29.5 m/s
The below section is not covered during lecture.
The speed of the source can exceed
the speed of the wave.
The envelope of these wave fronts is a
cone whose apex half-angle is given by
sin θ = vt / vsourcet
θ = sin-1(v / vsource)
(θ is the Mach angle)
The ratio vsource / v is referred to as the
Mach number
The conical wave front produced when
vsource > v is known as a supersonic “shock
wave”
The shock wave carries a great deal of
energy concentrated on the surface of
the cone
There are correspondingly great
pressure variations
Doppler Ball - OW-B-DB
14.7 Superposition and Interference
Differences between waves and particles
Particles have (1) zero size and (2) must exist at different location
Waves have (1) characteristic size, λ, and can (2) combine at one point
(a) If two or more traveling waves are moving through (c) For mechanical waves, linear waves have
a medium, the resultant value of the wave function at amplitudes much smaller than their wavelengths
any point is the algebraic sum of the values of the
wave functions of the individual waves
(d) Two traveling waves can pass through each other
without being destroyed or altered, because of the
(b) Waves that obey the superposition principle are
superposition principle
linear waves
Interference is the combination of separate waves in the same region of
space which produces a resultant wave
The two pulses, y1 & y2, with elements of positive
displacement are traveling in opposite directions with
same speeds
Waves start to overlap, the resultant wave function is
y1 + y2
When crest meets crest the resultant wave has a
larger amplitude than either of the original waves
The two pulses separate
They continue moving in their original directions The
shapes of the pulses remain unchanged
To the right
Two pulses are traveling in opposite directions
Their displacements are inverted with respect to each other
When they overlap, their displacements partially cancel each
other
Constructive Interference
When φ = 0 (crest to crest and trough
to trough), then cos (φ /2) = 1.
The amplitude of the
resultant wave is A1 + A2 = 2A.
The waves are “in phase.”
Destructive Interference
When φ = π (crest meets trough), then
cos (φ/2) = 0
and the amplitude of the resultant wave is 0
The wave are “out of phase.”
When φ is other than 0 or an even multiple of π,
the amplitude of the resultant is between 0 and
2A
The wave functions still add.
Sound from S can reach R by two different
paths.
The upper path can be varied
Constructive interference occurs @
Δr = |r2 – r1|
Δr = n λ
(n = 0, 1, …)
Destructive interference occurs @
Δr = |r2 – r1|
Δr = n ½λ
(n is odd)
φ occur  unequal path lengths
φ
=
Δr (2π λ)
Constructive interference
Δr = 2n λ/2
Δr = nλ
Destructive Inteference
Δr = (2n+1) λ/2
Δr = nλ + ½
Two waves are traveling in the same direction
along a stretched string. The waves are 90.0 out
of phase. Each wave has an amplitude of 4.00 cm.
Find the amplitude of the resultant wave.
Use the trig identity 
sina + sinb = 2[cos(a-b)/2][sin(a+b)/2]
y = 4 sin(kx – ωt) + 4 sin(kx – ωt + 90°)
Apply below trig identity
y = 2(4) cos(90°/2) sin(kx-ωt + 90°/2)
y = 8 cos(45°) sin (kx - ωt + 45°)
A = 8 cos(45°)
A = 5.66 cm
y = A sin (kx - ωt) + A sin (kx - ωt + φ)
y = 2A cos (φ /2) sin (kx - ωt + φ /2)
Destructive Interference Speakers: OW-B-Di
14.8 Standing Waves
Given two identical waves,
y1 = A sin (kx – ωt) and y2 = A sin (kx + ωt), they interfere according to the
superposition principle & with the trig identity: sin(a±b) = sina cosb ± cosa sinb
(Application Exercise: Apply the above identity)
we know y = y1 + y2 is
y = (2A sin kx) cos ωt
 This is the wave function of a standing wave
There is no kx – ωt term; therefore it is not a traveling wave, thus it’s called a standing wave
(appears to stands still)
There are three types of amplitudes used in describing waves
 The amplitude of the individual waves, A
 The amplitude of the simple harmonic motion of the elements in the medium,
 2A sin kx
 The amplitude of the standing wave, 2A
Nodes occurs at a point of zero amplitude
x = n λ /2
 n = 0, 1, 2, …
Antinode occurs at a point of maximum displacement, 2A
x = n λ /2 + λ /4
 n = 0, 1, 2, …
Recall: Resonance in Air Column from lab
Example
Two sinusoidal waves traveling in opposite
directions interfere to produce a standing wave
with the wave function, y = (1.50 m) sin(0.400x)
cos(200t), where x is in meters and t is in
seconds. Determine the wavelength, frequency,
and speed of the interfering waves.
k = 2 π /
0.4 = 2 π /
=5π
 = 15.7 meters
v= f 
v = 31.8 (15.7)
v = 500 m/s
ω = 2π f
200 = 2 π f
f = 100/ π
f = 31.8 Hz
to
apply trig identity
sin(a±b) = sina cosb ± cosa sinb
(a = kx
and b = ωt)
y = A sin (kx - ωt) + A sin (kx + ωt)
sin(a±b) = sina cosb ± cosa sinb
sin (kx - ωt) = sin kx cos ωt - cos kx sin ωt
sin (kx + ωt) = sin kx cos ωt + cos kx sin ωt
sin (kx - ωt) + sin (kx + ωt) = 2 sin kx cos ωt
y = (1.50 m) sin(0.400x) cos(200t)
Standing Waves, Cenco String: OW-B-Sc
Standing Waves in a String Fixed at Both Ends
Consider a string fixed at both ends
 The string has length L
 Standing waves are set up by a continuous superposition of waves incident on
and reflected from the ends
 There is a boundary condition on the waves
 The wavelengths of the normal modes for a string of
length L fixed at both ends are λ= 2L / n
n = 1, 2, 3, …
 n is the nth normal mode of oscillation
 These are the possible modes for the string
The natural frequencies are
fnat = n
v / 2L
fnat = n √(FT/μ)/ 2L
The above situation in which only certain
frequencies of oscillation are allowed is called
quantization
Quantization is a common occurrence when waves are subject
to boundary conditions
The fundamental frequency occurs at n = 1,
and is the lowest frequency, ƒ1
Harmonics are integral multiples or the
fundamental frequency and together they
form the harmonic series
A musical note is defined by its
fundamental frequency, (n = 1)
Example:
The frequency of the string can be changed by
changing either its length or its tension (changing
it’s boundary condition)
Example
Find the fundamental frequency and the
next three frequencies that could cause
standing-wave patterns on a string that
is 30.0 m long, has a mass per length of
9.00 x 10–3 kg/m, and is stretched to a
tension of 20.0 N.
The above string is vibrating in its second harmonic
A middle “C” on a piano has a
fundamental frequency of 262
Hz. What are the next two
harmonics of this string?
v2 = FT / μ
v2 = 20 / .009
v = 47.1 m/s
fnat = n v/2L
f1 = n v/2L
f1 = 47.1/(2*30)
f1 = 0.786 Hz
Ans:
ƒ1 = 262 Hz
ƒ2 = 2ƒ1 = 524 Hz
ƒ3 = 3ƒ1 = 786 Hz
f2 = 2(47.1)/(2*30)
f2 = 1.57 Hz
f3 = 3(47.1)/(2*30)
f3 = 2.36 Hz
f4 = 4(47.1)/(2*30)
f4 = 3.14 Hz
Resonance
Resonance can occur in systems that are capable of oscillating in one or more normal modes
If a periodic force is applied to such a system, the
amplitude of the resulting motion is greatest when the
frequency of the applied force is equal to one of the
natural frequencies of the system
Because an oscillating system exhibits a large amplitude
when driven at any of its natural frequencies, these
frequencies are referred to as resonance frequencies,
ƒo = largest amplitude
The maximum amplitude is limited by friction in the
system (of the medium the wave is traveling)
Example
The chains suspending a child's swing are 2.00 m long. At
what frequency should a big brother push to make the child
swing with largest amplitude?
Standing Waves in Air Columns
ω = √(g/L)
2πf = √(g/L)
2πf = √(9.8/2)
f = 0.352
The closed end of a pipe is a
displacement node because the wall at
this end doesn’t allow longitudinal motion Closed one end
of the air molecules. Thus the reflected fnat = n v/4L
wave is 180° out of phase with the
v = f1 λ
incident wave.
Also the pressure wave is 90° out of
phase with the displacement wave, 17.2, Open both ends
we know the closed end of the tube
fnat = n v/2L
corresponds to a pressure ANTINODE.
The open end is a displacement AND
pressure node.
The overall length of a piccolo is 32.0 cm. The resonating
air column vibrates as in a pipe open at both ends. (a) Find
the frequency of the lowest note that a piccolo can play,
assuming that the speed of sound in air is 340 m/s. (b)
Opening holes in the side effectively shortens the length of
the resonant column. If the highest note a piccolo can
sound is 4000 Hz, find the distance between adjacent
antinodes for this mode of vibration.
Extra … Standing Waves in Rods
dAtoA = 0.320 m
L = λ /2
λ = 0.32 (2)
λ = 0.64 meters
v =f λ
340 = f 0.64
f = 531 Hz
(b)
v =f λ
340 = 4000 λ
λ = 0.0850 m
dAtoA = 0.0425 m
If a rod is clamped at a node, then a standing wave may be set up in a rod. Just add energy by
rubbing rosin pad or similar material along the length of the unclamped portion of the rod.
Singing Rods: OW-D-Sr
An aluminum rod is clamped one quarter of the way
along its length and set into longitudinal vibration
by a variable-frequency driving source. The lowest
frequency that produces resonance is 4400 Hz.
The speed of sound in an aluminum rod is 5100 m/s.
Determine the length of the rod.
14.9
When the rod is clamped at one-quarter of its
length, the vibration pattern reads ANANA and
the rod length is L = 2dAtoA = λ.
v = f L  v =f λ
5100 = 400(L)
L = 1.16 meters
Beats (Interference in Time)
Beating is the periodic
variation in amplitude at a
y1 = A cos ω1t
y1 = A cos 2πf1t
y2 = A cos ω2t
y2 = A cos 2πf2t
given point due to the
superposition of two waves
having slightly different
frequencies.
where Anew = 2Acos((πt(f1 - f2)) 
cos a
+ cos b = 2cos((a-b)/2)cos((a+b)/2)
y = y1 + y2
y = A cos 2πf1t + A cos 2πf2t
apply the above trig identity
y = 2Acos((2πf1t - 2πf2t)/2) cos((2πf1t + 2πf2t)/2)
y = 2Acos((2πt(f1 - f2)/2)
cos((2πt(f1 + f2)/2)
y=
Anew
cos((2πt(f1 + f2)/2)
The above derivation isn’t explained in the book…only the result.
And cos((2πt(f1 - f2)/2) = ± 1
Since amplitude varies with ½(f1 - f2) we get two max amplitudes during one period
So we get a beat  fbeat = |f1 – f2|
In certain ranges of a piano keyboard, more than
one string is tuned to the same note to provide
extra loudness. For example, the note at 110 Hz
has two strings at this frequency. If one string
slips from its normal tension of 600 N to 540 N,
what beat frequency is heard when the hammer
strikes the two strings simultaneously?
fλ = v = (FT/μ)1/2
fnew / f = (FT-new/ FT)1/2
fnew / 110 = (540/ 600)1/2
fnew = 104.4 Hz
Δf = 110 – 104.4
Δf = 5.64 beats/sec
Differential Tuning Forks: OW-B-Df
14.11 Rate of Energy Transfer by Sinusoidal Waves on Strings (Optional)
ΔK = ½ m v2
ΔK = ½(μΔx)v2
dK = ½ μ ω2A2 cos2(kx) dx
∫dK = ½ μ ω2A2 ∫cos2(kx) dx
So our total energy is
TE = ¼ μ ω2A2 λ + ¼ μ ω2A2 λ
μ is mass per unit length
v is the transverse speed
integrate by parts
K = ½ μ ω2A2 [x/2+sin(2kx)/4k]
dK = ½ μ dx [ v2 ]
dK = ½ μ dx [ω2A2 cos2(kx – ωt)]
dK = ½ μ ω2A2 cos2(kx – ωt) dx
@t=0
(ωt just shifts the phase, thus if
Example
A taut rope has a mass of 0.180 kg and a length
of 3.60 m. What power must be supplied to the
rope in order to generate sinusoidal waves having
an amplitude of 0.100 m and a wavelength of
0.500 m and traveling with a speed of 30.0 m/s?
The Linear Wave Equations
ΣFy
ΣFy
m
ay
m (∂2y/∂t2)
μΔx (∂2y/∂t2)
= FT sinθB
= FT (∂y/∂xB)
= FT (∂y/∂xB)
= FT (∂y/∂xB)
= FT [∂y/∂x B
–
–
–
–
–
FT sinθA
FT (∂y/∂x A)
FT (∂y/∂x A)
FT (∂y/∂x A)
∂y/∂xA]
(μ/ FT) ∂2y/∂t2 = [∂y/∂xB – ∂y/∂xA] / Δx
Work = ΔK
TE = ΔK for one full period
2
Using similar derivation from
above, we get our Potential energy
function as
U = ¼ μ ω2A2 λ
we’re attempting to find Total
Energy a phase shift is irrelevant)
14
from 0 to λ
K=¼μω A λ
2
TE = ½ μ ω2A2 λ
P = Work / time
P = ΔK / T
P = ½ μ ω2A2 λ / T
v = f λ = λ/T
P = ½ μ ω2A2 v
v
=
fλ
30.0 = f (0.500)
f =60.0 Hz
ω = 2πf
ω = 2π(60)
ω = 120π rad/s
v
P =½(0.18/3.6) (120π)2 (0.100)2 30.0
P = 1070 Watts
P=½
(optional)
(eq 16.23)
(eq 16.24)
(eq 16.25)
At this point we need to remember that at some point in time, tension at B
was or will be the same as the tension at A at the current time, so that
μ
ω2
A2
Tsin(θB + ωt) - TsinθA = 0 at some time, t.
Notes & explanations
The partial derivative of any function is
Associate f(x + Δx) with ∂y/∂xB (1st derivative of point B) and f(x)
with ∂y/∂xA (1st derivative of point A)
 The derivative of a point is the slope (1st derivative), and the definition
of a partial derivative gives us the 2nd derivative, thus ∂2y/∂x2 
(μ/T) ∂2y/∂t2
= ∂2y/∂x2
This is the linear wave equation for a string.
y(x, t) = A sin (kx – ωt) (from 16.2)
∂2y/∂t2
= (FT /μ) ∂2y/∂x2
∂2y/∂t2
y(x, t) = A sin (kx – ωt)
∂y/∂t = -ω A cos (kx – ωt)]
∂2y/∂t2 = -ω2 A sin (kx – ωt)]
      ∂2y/∂t2
-ω2 A sin(kx – ωt)]
(sinθ)max = (cosθ)max = 1
-ω2 A
ω2 / k2
v2
T represents the tension which is a force, we
should say FTension.
ΣFy = TsinθB – TsinθA
(TB is going up…thus
positive)
If θ is small, θ < 10° we know
(always applicable at small amplitudes)
sin θ = tan θ;
&
tan θ = y / x;
And since our infinitesimal displacements
(sin θ = dy/dx) are changing with time
(& only concerned with components of “y” that
are dependant on “x”)
sin θ = ∂y/∂x
=
∂2y/∂x2
y(x, t) = A sin (kx – ωt)
∂y / ∂x = k A cos (kx – ωt)]
∂2y/∂x2 = -k2 A sin (kx – ωt)]
(FT /μ)
∂2y/∂x2
= (FT /μ) -k2A sin(kx – ωt)]
= (FT /μ) -k2A
= FT / μ
= FT / μ
     
∂2y/∂t2 = (FT /μ) ∂2y/∂x2
∂2y/∂x2 = (μ/ FT) ∂2y/∂t2
∂2y/∂x2 = (1/v2) ∂2y/∂t2
Example
Show that the wave function
y = ln[b(x – vt)] is a solution to
∂2y/∂x2 = (1/v2) ∂2y/∂t2, where b is a
constant.
y
= ln[b(x – vt)]
∂y/∂t = -bv / b(x-vt)
∂2y/∂t2 = -1(-bv)(bv) / b2(x-vt)2
∂2y/∂t2 = v2 / (x-vt)2
(1/v2) ∂2y/∂t2
=
Yes, it satisfies the equation
y
= ln[b(x – vt)]
∂y/∂x = -b / b(x-vt)
∂y/∂x = -1 / (x-vt)
∂2y/∂x2 = -1 / (x-vt)2
∂2y/∂x2