Week 10 Waves Ch15 16 FINAL
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Transcript of Week 10 Waves Ch15 16 FINAL
Week 10 Chapter 15 & 16 Waves and Oscillations
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Lecture 1:
Oscillations
- types of waves 16.2
- transverse and longitudinal waves 16.3
- wavelength, frequency and speed of wave 16.4, 16.5
Lecture 2:
Waves
- simple harmonic motion 15.2, 15.3
Lecture 3:
Waves
- wave speed on stretched string 16.6
- energy and power 16.7
Week 10
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• A wave is an oscillation that moves through space or matter
transferring energy from one place to another
• Generally – material returns to its original position as the wave
passes by –there for there is no mass transport.
Waves
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Classical waves transfer energy without transporting matter through the
medium. Waves in a medium do not move the material in the medium
from one place to another place; instead the wave's energy travels
through the medium, leaving the material in place, similar to a cork
rising and falling in one place as the wave moves past the cork
cork bobs up and
down at the same
position
Waves – Transfer of Energy
16.2 Types of Waves
• Mechanical waves. These waves have two central features: They
are governed by Newton’s laws, and they can exist only within a
material medium – NO medium NO wave: water waves, sound
waves, and seismic waves (earthquake).
• Electromagnetic waves. These waves require no material medium
to exist = PURE ENERGY. All electromagnetic waves travel
through a vacuum at the same exact speed c, speed of light = 3 x108
ms-1. Common examples include visible and ultraviolet light, radio
and television waves, microwaves, x rays, and radar (radio) waves.
• Matter waves. These waves are associated with electrons, protons,
and other fundamental particles, atoms and molecules moving at
very high speeds where they acquire wave-like properties. Basis of
quantum mechanics. Louise de Broglie (1882 – 1987) λ =ℎ
𝑚𝑣.
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In a transverse wave, the
displacement of every such oscillating
element along the wave is
perpendicular to the direction of travel
of the wave
16.3 Transverse and Longitudinal Waves
In a longitudinal wave the motion of
the oscillating particles is parallel to
the direction of the wave’s travel
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Note – energy of wave inversely
proportional to its energy 𝐸 ∝1
𝜆
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l
A
Time/distance dis
pla
cem
ent
Wavelength, l, distance between two crests or troughs –
distance to complete one cycle
Amplitude – maximum displacement
Period, T – time between 2 crests/troughs – time to complete
one cycle
Frequency, f = 1/T in Hertz (Hz) – number of oscillations per second
vs
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Velocity of a wave, vs = distance/time = l/T = l.f
Waves
q
l
A
Time/distance
dis
pla
cem
ent 1l = 1 cycle = 360 = 2p
y = Asin(q)=Acos(p/2 – q)
y y A
Angular frequency w = 2p.f angular speed, number of cycles per sec, in
radians per sec = rad.s-1
1 𝑟𝑎𝑑 = 1800
𝜋 1 =
𝜋
1800
Angular wave number, k = 2π/λ- number of cycles per unit distance
the spatial equivalent to frequency = “spatial frequency” of the wave
k = phase change of the traveling wave in terms of rad.m-1
vs
k = w/v = proportionality between w and v
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16.4 Wavelength and Frequency Wave Motion
1s
angular frequency = number of cycles per sec = 2𝜋
𝑇 = 2pf rad.s-1
1m
wavenumber = number of cycles per metre = 2𝜋
𝜆 rad.m-1
T
l
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Handbook of Recording
Engineering, 1986, p. 2
16.4 Wavelength and Frequency Wave Motion
y = A.sin.f
f
Phase, f, relationship between
2 waves with same frequency
k = number of 2p cycles per
unit distance
Each full cycle f = 360
Note: height is the same whenever x is
an integral multiple of l
Phase angle at distance, x
f = k.x = (2p/l).x
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Handbook of Recording
Engineering, 1986, p. 2
16.4 Wavelength and Frequency Wave Motion
y = A.sin.f
f
Dx
y = Asin2π
λ. x = Asin k. x
x xt
w = k.v
16.4 Wavelength and Frequency Wave Motion
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360 = 2p
f y
y = A.sin.f
l
A
x dis
pla
cem
ent
If the wave is moving at velocity, v, it moves a distance xt =vt
A
w = k.v
y = Asin2π
λ. x = Asin k. x
16.4 Wavelength and Frequency –wave motion
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y(x,t) = ym sin (kx ± wt –fo )
General Form
Phase Constant (radians)
- wt wave traveling to right
+ wt wave traveling to left
used to represent the wave as a function of time, t, at a specific distance, x,
or as a function of distance, x, at a specific time, t.
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When 2 sine waves of equal amplitude and frequency propagate
through a medium in opposite directions – combine to form a
Standing Wave. Note: No Energy Transfer with Standing Waves
Wave moving to the right is: A sin(kx - ωt) & the left is: A sin(kx + ωt)
Standing wave – addition of the 2 waves
y(x,t) = A sin(kx - ωt) + A sin(kx + ωt)
Using sin A+sin B = 2sin [(A+B)/2] .cos [(A-B)/2]
y ( x, t ) = 2A cos(ωt) * sin kx
Where 2A cos(ωt) determines how the amplitude varies with time
While sin(kx) determines the standing wave’s shape.
Standing Waves
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Standing Waves
Fundamental frequency
½ sine waveform = ½ l
2nd Harmonic
1 sine waveform = 1 l
3rd Harmonic
3/2 sine waveform = 3/2 l
4th Harmonic
2 sine waveform = 2 l
l = 2𝐿
𝑛 where n = 1, 2, 3, .......
n = number of nodes
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Standing Waves Examples
atomic
orbitals
standing wave
solution for
electrons in an
atom
16.4 Wavelength and Frequency
The amplitude ym of a wave is the magnitude
of the maximum displacement of the elements
from their equilibrium positions as the
wave passes through them.
The phase of the wave is the argument
(kx –wt) of the sine function. As the wave
sweeps through a string element at a particular
position x, the phase changes linearly with
time t.
The wavelength l of a wave is the distance
parallel to the direction of the wave’s
travel) between repetitions of the shape of the
wave (or wave shape). It is related to the
angular wave number, k, by
:
The period of oscillation T of a wave is the
time for an element to move through one full
oscillation. It is related to the angular
frequency, w, by
The frequency f of a wave is defined as 1/T
and is related to the angular frequency w by
A phase constant f in the wave function:
y =ym sin(kx –wt+ f). The value of f can be
chosen so that the function gives some other
displacement and slope at x 0 when t 0.
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16.4 Wavelength and Frequency: The Speed of a Traveling Wave
As the wave in Fig. 16-7 moves,
each point of the moving wave
form, such as point A marked on a
peak, retains its displacement y.
(Points on the string do not retain
their displacement, but points on the
wave form do.) If point A retains its
displacement as it moves, the phase
giving it that displacement must
remain a constant:
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Example, Transverse Wave
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Example, Transverse Wave, Transverse Velocity, and Acceleration
velocity 𝑥, 𝑡 = 𝑣𝑦 = 𝑑𝑦
𝑑𝑡 = −𝜔𝑦𝑚cos 𝑘𝑥 − 𝜔𝑡
y 𝑥, 𝑡 = 𝑦𝑚sin 𝑘𝑥 − 𝜔𝑡
acceleration 𝑥, 𝑡 = 𝑎𝑦= 𝑑𝑣
𝑑𝑡 = −𝜔2𝑦𝑚sin 𝑘𝑥 − 𝜔𝑡
y, vy & ay vs
vy,m = -wym
ay,m = -w2ym
transverse wave speed
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y = Asinf v = -wAcosf
𝑠𝑖𝑛𝜙 =𝑦
𝐴 𝑐𝑜𝑠𝜙 = −
𝑣
𝜔𝐴
𝑠𝑖𝑛2𝜙 + 𝑐𝑜𝑠2𝜙 = 1 =𝑦2
𝐴2 +
𝑣2
𝜔2𝐴2
𝑣2=𝜔2 𝐴2 − 𝑦2
Relationship between v, w, A and y
Example, Transverse Wave, Transverse Velocity, and Acceleration
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Chapter 15
Oscillations
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15.1 Oscillatory motion
Motion which is periodic in time, that is, motion that
repeats itself in time.
Examples:
• Power line oscillates when the wind blows past it
• Earthquake oscillations move buildings
• Weight oscillating on a spring
• Tuning Fork
• Pendulum
• Piston moving in a car engine
• Guitar String
Sometimes the oscillations are so severe, that the
system exhibiting oscillations break apart. 24
15.2 Simple Harmonic Motion
• A particle repeatedly moves back and forth about the point x=0 where the
time taken for 1 complete oscillation is the period, T.
• In T the particle travels from x=+xm, to –xm, and then back to its original
position xm.
• The velocity vector arrows are scaled to indicate the magnitude of the
speed of the system at different times. At x=±xm, the velocity is zero. 25
mg
m
pendulum
Frequency of oscillation is the number of oscillations that
are completed in each second.
The symbol for frequency is f, and the SI unit is the Hertz
(abbreviated as Hz).
It follows that (sec)
1)(
THzf =
15.2 Simple Harmonic Motion
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Any motion that repeats itself is periodic (= harmonic).
If the motion is a sinusoidal function of time, it is called
simple harmonic motion (SHM).
Mathematically SHM can be expressed as:
)cos()( fw = txtx m
Here,
• xm is the amplitude (maximum displacement of the system)
• t is the time
• w is the angular frequency, and
• f is the phase constant or phase angle
15.2 Simple Harmonic Motion
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(a) The displacement of two SHM
systems that are different in
amplitudes, but have the same period.
(b) The displacement of two SHM
systems which are different in periods
but have the same amplitude.
(c) The value of the phase constant
term, f, depends on the value of the
displacement and the velocity of the
system at time t = 0. The
displacement of two SHM systems
having the same period and
amplitude, but different phase
constants.
15.2 Simple Harmonic Motion
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(a)
(c)
(b)
For an oscillatory motion with period T,
)()( Ttxtx =
The cosine function also repeats itself when the argument
increases by 2p. Therefore,
fT
T
tTt
pp
w
pw
pww
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2
2)(
==
=
=
Here, w is the angular frequency, and measures the
angle per unit time. Its SI unit is radians/second. To be
consistent, f must also be in radians.
15.2 Simple Harmonic Motion
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15.2 Simple Harmonic Motion
The velocity of SHM
)sin()(
cos()(
)(
fww
fw
=
==
txtv
txdt
d
dt
tdxtv
m
m
The maximum value of velocity is: v = wxm.
The phase shift of the velocity is p/2, making
the cosine to a sine function. The acceleration
of SHM is:
)()(
)cos()(
)sin()(
)(
2
2
txta
txta
txdt
d
dt
tdvta
m
m
w
fww
fww
=
=
==
The acceleration amplitude is: a = - w2xm.
In SHM a(t) is proportional to the displacement but opposite in sign. 30
a(t) maximum where
maximum F applied
a(t) maximum
at bottom of
swing.
15.3 Force Law for SHM
From Newton’s 2nd law:
kxxmmaF === 2w
SHM is the motion executed by a system subject to a force that is
proportional to the displacement of the system but opposite in sign.
Period T of SHM oscillation =
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F = -kx
Hooke’s law
k is the spring constant
m
k=w
22 )2( fmmk pw ==
k
mT p2=
m
kf
p2
1=
15.3 The Force Law for Simple Harmonic Motion
The block-spring system shown on
the right forms a linear SHM
oscillator.
The spring constant of the spring,
k, is related to the angular
frequency, w, of the oscillator:
k
mT
m
kpw 2==
32 mg
m
pendulum
T = 2p𝐿
𝑔
L
Simple pendulum
• Perpendicular to the string
mg sin q = -ma
a is the acceleration along the arc
sin q ≈ q and x ≈ Lq ,
mg sin q = -ma = mg q
a = -g(x/l) = -(g/l)x
as a = -w2x in SHM then w2 = g/l
• Period of oscillation T
g
L2
2T p=
w
p=
A
F=mg
mg cos q
mg sin q
q
x
T
L
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Combination of Springs
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F
F2 = k2x F1 = k1x
F = k1x + k2x = (k1 + k2) x
F
F1 = k1x1
F2 = k2 x2
==
21 k
1
k
1
xxkF
𝟏
𝒌 =
𝟏
𝒌𝟏+𝟏
𝒌𝟐
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Combination of Springs
F = T1 + T2 = k1x + k2x = (k1 + k2)x
T1
Equilibrium position
T2
compression
of spring
extension
of spring
Spring
constant, k1
Spring
constant, k2
x
motion of block
Example: Force law:
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Example, force law:
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Example, force law:
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Example, force law:
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Example, force law:
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Example, force law:
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15.4: Energy in Simple Harmonic Motion (not in exam)
The potential energy of a linear oscillator
is associated entirely with the spring.
fw == tkxkxtU m
222 cos2
1
2
1)(
The kinetic energy of the system is
associated entirely with the speed of
the block.
fwfww === tkxtxmmvtK mm
222222 sin2
1sin
2
1
2
1)(
The total mechanical energy of
the system:
222
m kx2
1mv
2
1kx
2
1KUE ===
42 energy oscillates between KE and PE
Example, energy in SHM:
Many tall building have mass dampers,
which are anti-sway devices to prevent
them from oscillating in a wind. The
device might be a block oscillating at the
end of a spring and on a lubricated track.
If the building sways, say eastward, the
block also moves eastward but delayed
enough so that when it finally moves, the
building is then moving back westward.
Thus, the motion of the oscillator is out of
step with the motion of the building.
Suppose that the block has mass m = 2.72
x 105 kg and is designed to oscillate at
frequency f = 10.0 Hz and with amplitude
xm = 20.0 cm.
(a) What is the total mechanical energy E
of the spring-block system?
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Example, energy, continued:
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16.6: Wave Speed on a Stretched String
The speed of a wave, v , along a stretched ideal string
depends only on the tension, t , and linear density of the
string, m , and not on the frequency, f , of the wave.
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m Wave
driver t = mg N
𝜇 =𝑚𝑠𝑡𝑟𝑖𝑛𝑔
𝑙𝑠𝑡𝑟𝑖𝑛𝑔 𝑣 = 𝜏
𝜇
Dl
vspeed
16.6: Wave Speed on a Stretched String
A small string element of length Dl
within the pulse is an arc of a circle
of radius R and subtending an angle
2q at the center of that circle.
A force with a magnitude equal to
the tension in the string, t, pulls
tangentially on this element at each
end
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t cosq t cosq
t sinq t sinq
tangential q
𝑠𝑖𝑛𝜃 =Δ𝑙2
𝑅 = q
16.6: Wave Speed on a Stretched String The horizontal components of these forces
cancel, but the vertical components add to form
a radial restoring force . For small angles,
If m is the linear mass density of the string, and
Dm the mass of the small element,
The element has an acceleration:
Therefore, as F = ma
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t cosq t cosq
t sinq t sinq
tangential q
𝑠𝑖𝑛𝜃 =Δ𝑙2
𝑅 = q
𝑣 =𝜔
𝑘= 𝜏
𝜇 𝜇 = 𝜏
𝑣2
𝑓 =1
𝜆𝜏𝜇 𝜏𝑘2=𝜇𝜔2
16.6: Wave Speed on a Stretched String
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R R
Velocity, t2 = v2
Velocity, t1 = v1 Distance
travelled
t2 >t1
Change in
velocity
q
q Dv = v2 –v1
𝑑𝑖𝑠𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑
𝑟= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑣.Δ𝑡
𝑟 =
Δ𝑣
𝑣
Δ𝑣
Δ𝑡 = a =
𝑣2
𝑟
Since v1 and v2 are perpendicular to R, the angles of both
triangles are equal. Consequently the two triangles are
similar, the ratios of corresponding sides are equal.
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16.6: Wave Speed on a Stretched String
Vibration frequency of a stretched string depends on:
• The length of the string that is free to oscillate - longer
string lower frequency
• The tension in the string - higher tension higher frequency
• The mass of the string - higher mass lower frequency
𝑓 =1
𝜆 𝜏
𝜇 =
𝑛
2𝑙
𝜏 . 𝐿
𝑚 𝑓 =
𝑛
2
𝜏
𝑚𝐿 l =
2𝐿
𝑛
16.6: Energy and Power of a Wave Traveling along a String
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Kinetic energy of string element
at each position depends on the
transverse velocity of the element
Potential energy depends on the
amount by which the string element
is stretched as the wave passes
through it
Snapshot of a travelling wave on a
string at t = 0
l
T
Energy, El
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• In one period T, energy has moved one wavelength, l
• Transported Power, P = 𝐸𝜆
𝑇
𝑷 =𝟏
𝟐 𝝁𝝕𝟐𝑨𝟐
𝝀
𝑻=𝟏
𝟐 𝝁𝝕𝟐𝑨𝟐𝒗
16.6: Energy and Power of a Wave Traveling along a String
𝑬𝝀 = 𝟏
𝟐 𝝁𝝕𝟐𝑨𝟐 𝝀
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16.6: Energy and Power of a Wave Traveling along a String
𝑷𝝀 =𝟏
𝟐 𝝁𝒘𝟐𝑨𝟐𝒗𝒔𝒑𝒆𝒆𝒅
Mass per unit length
of the string (kg/m)
Angular frequency of
wave (radians/sec)
wave velocity
speed (m/s)
Amplitude of wave (m)
Power (Watts)
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16.7: Energy and Power of a Wave Traveling along a String
l
vs
Wave motion
A y
vy
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 = ∆𝐾𝐸 = 1
2𝑚𝑣𝑦
2 = 1
2𝜇 Δ𝑥 𝑣𝑦
2
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = ∆𝑈 =1
2𝐹. 𝑦 =
1
2(𝑚. 𝑎). 𝑦 =
1
2𝑚𝜔2𝑦2
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = ∆𝑈 = 1
2𝑚𝜔2𝑦2 =
1
2 m Dx 𝜔2𝑦2
𝜇 =𝑚
Δ𝑥
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16.6: Energy and Power of a Wave Traveling along a String
y = Asin(kx – wt)
∆𝑈 = 1
2𝑚𝜔2𝑦2 =
1
2𝜔2𝑦2mDx =
1
2m𝜔2A2sin2(kx-wt)Dx
Wave equation – travelling wave
Potential energy for one wavelength along the string is found by
integrating over dx at t=0
𝑈𝜆 =1
2 𝜇𝜔2𝐴2 𝑠𝑖𝑛2 𝑘𝑥 𝑑𝑥 =
1
2 𝜇𝜔2𝐴2
1
2𝑥 −
1
4𝑘𝑠𝑖𝑛 2𝑘𝑥
0
𝜆𝜆
0
= 1
4𝜇𝜔2𝐴2𝜆
Total Energy with one wavelength = El = Ul + KEl = 1
2𝜇𝜔2𝐴2𝜆
Δ𝐾𝐸 =1
2𝑚𝑣2 =
1
2𝜇𝑣2Δ𝑥 v = wAcos(kx–wt)
∆𝐾𝐸 = 1
2m𝜔2A2cos2(kx-wt)Dx 𝐾𝐸𝜆 =
1
2 𝜇𝜔2𝐴2 𝑐𝑜𝑠𝑑𝑥
𝜆
0
= 1
4𝜇𝜔2𝐴2𝜆
Similarly
𝑠𝑖𝑛2 𝑘𝑥 = 1
21 − 𝑐𝑜𝑠 𝑘𝑥
16.6: Energy and Power of a Wave Traveling along a String
The average power, which is the average rate at which energy of both kinds
(kinetic energy and elastic potential energy) is transmitted by the wave, is:
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Velocity of string Kinetic energy of string
ym = A
Example, Transverse Wave:
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57
A string of mass of 20 mg and length of 1000 mm carries a wave
with a period 65 ms and a wavelength 60 cm. What is the average
power transmitted along the string if the waves amplitude is 12
mm?
𝑃𝜆 =1
2 𝜇𝑤2𝐴2𝑣
v = l .T -1 w = 2p .T -1
𝑃𝜆(𝑊𝑎𝑡𝑡) =1
2 𝜇4𝜋2
𝑇2𝐴2𝜆
𝑇
Here A = 12 x10-3 m, l = 0.6 m, T = 6.5 x10-2 s, μ = 2 x10-3 kg/m
P = 12.42 Watts