Wave Motion - University of Louisville
Transcript of Wave Motion - University of Louisville
1Prof. Sergio B. MendesSummer 2018
Chapter 14 of Essential University Physics, Richard Wolfson, 3rd Edition
Wave Motion
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Waves: propagation of energy, not particles
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Longitudinal Waves:disturbance is along the direction of
wave propagation
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Transverse Waves:disturbance is perpendicular to the
direction of wave propagation
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Waves with Longitudinal Transverse Motions
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Amplitude of a Wave
height
pressure
longitudinal displacement
transverse displacement
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a pulse
a wave train
a continuous wave
Waveforms
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PhET
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Wavelength in a continuous wave
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Wave Speed
π£π£ =ππππ
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PhET
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Two Snapshots of a Wave Pulse
π‘π‘ = 0
π¦π¦ = ππ π₯π₯
π‘π‘ β₯ 0
π¦π¦ = ππ π₯π₯ β π£π£ π‘π‘
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PhET
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ππ(π₯π₯, π‘π‘) = ππ π₯π₯ β π£π£ π‘π‘
Fingerprint of a Wave:
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A Harmonic Wave
ππ π₯π₯, π‘π‘ = 0 = π΄π΄ ππππππ 2 πππ₯π₯ππ
ππ π₯π₯, π‘π‘ = π΄π΄ ππππππ 2 πππ₯π₯ β π£π£ π‘π‘ππ
π‘π‘ = 0
π‘π‘ β₯ 0
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PhET
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A Couple of Definitions
ππ β‘2 ππππ
Wave Number
ππ π₯π₯, π‘π‘ = π΄π΄ ππππππ 2 πππ₯π₯ β π£π£ π‘π‘ππ
ππ β‘2 ππππ
Angular Frequency
= 2 ππ ππ
ππ π₯π₯, π‘π‘ = π΄π΄ ππππππ ππ π₯π₯ β ππ π‘π‘
π£π£ =ππππ
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Propagation towards Positive x-directionππ π₯π₯, π‘π‘ = π΄π΄ ππππππ ππ π₯π₯ β ππ π‘π‘
ππ π₯π₯, π‘π‘ = π΄π΄ ππππππ ππ π₯π₯ + ππ π‘π‘
Propagation towards Negative x-direction
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Got It ? 14.1
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The Wave Equationππ(π₯π₯, π‘π‘) = π΄π΄ ππππππ ππ π₯π₯ β ππ π‘π‘
πππππππ₯π₯
= β ππ π΄π΄ πππ π π π ππ π₯π₯ β ππ π‘π‘
ππ2πππππ₯π₯2
= β ππ2 π΄π΄ ππππππ ππ π₯π₯ β ππ π‘π‘
πππππππ‘π‘
= ππ π΄π΄ πππ π π π ππ π₯π₯ β ππ π‘π‘
ππ2πππππ‘π‘2
= β ππ2 π΄π΄ ππππππ ππ π₯π₯ β ππ π‘π‘
1ππ2
ππ2πππππ₯π₯2
=1ππ2
ππ2πππππ‘π‘2
ππ2πππππ₯π₯2
=1π£π£2
ππ2πππππ‘π‘2
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Waves on a String
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An example on how the properties of the carrying medium determines
the wave speed:
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πΉπΉππππππ β 2 πΉπΉ ππ
ππ β 2 ππ π π ππ
ππ =π£π£2
π π 2 πΉπΉ ππ β 2 ππ π π ππ
π£π£2
π π
π£π£ =πΉπΉππ
πΉπΉ = ππ ππ
Wave Speed
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Example 14.2
ππ = 5.0 ππππ
βπ₯π₯ = 43 ππ
βπ‘π‘ = 1.4 ππ
π£π£ =πΉπΉππ
πΉπΉ = ? ?
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Got It ? 14.2
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Wave Power
ππ = ππ.ππ
= βπΉπΉ π£π£ πππ π π π ππ
β βπΉπΉ π£π£ π‘π‘πππ π ππ
π¦π¦ π₯π₯, π‘π‘ = π΄π΄ ππππππ ππ π₯π₯ β ππ π‘π‘
π£π£ =πππ¦π¦πππ‘π‘
π‘π‘πππ π ππ =πππ¦π¦πππ₯π₯
= ππ π΄π΄ πππ π π π ππ π₯π₯ β ππ π‘π‘
= βππ π΄π΄ πππ π π π ππ π₯π₯ β ππ π‘π‘
= πΉπΉ ππ ππ π΄π΄2 πππ π π π 2 ππ π₯π₯ β ππ π‘π‘
= ππ π£π£ ππ2 π΄π΄2 πππ π π π 2 ππ π₯π₯ β ππ π‘π‘
ππ = ππ π£π£ ππ2 π΄π΄2 πππ π π π 2 ππ π₯π₯ β ππ π‘π‘οΏ½ππ =12ππ π£π£ ππ2 π΄π΄2
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Wave Intensity
πΌπΌ β‘πππππππππππ΄π΄ππππππ
πΌπΌ =ππππππππππ4 ππ ππ2
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Example of Wave Intensities
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Example 14.3ππ1 = 9.2 ππ
π₯π₯1 = 1.9 ππ π₯π₯2 = ? ?
ππ2 = 4.9 ππ
πΌπΌ =ππππππππππ4 ππ ππ2
πΌπΌ1 =ππ1
4 ππ π₯π₯12
πΌπΌ2 =ππ2
4 ππ π₯π₯22πΌπΌ1 = πΌπΌ2
ππ14 ππ π₯π₯12
=ππ2
4 ππ π₯π₯22π₯π₯2 = π₯π₯1
ππ2ππ1
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Got It ? 14.3
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Sound Waves
π£π£ =πΎπΎ ππππ
πΎπΎ is a constant characteristic of the gas
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π½π½ ππππ β‘ 10 log10πΌπΌπΌπΌππ
πΌπΌππ β‘ 1 Γ 10β12 ππππ2
Audible Frequencies for Human Ears
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Example 14.4π½π½1 = 75 ππππ
ππ2ππ1
= ? ?π½π½2 = 60 ππππ
ππ2ππ1
=πΌπΌ2πΌπΌ1
πΌπΌ2πΌπΌ1
=πΌπΌππ 10
π½π½210
πΌπΌππ 10π½π½110
= 10π½π½2βπ½π½110
πΌπΌ = πΌπΌππ 10π½π½10
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Interferenceor what happened when two waves are present in the same region of space at a particular
time ?
Just add them up !!
β’ When wave crests coincide with crests, the interference is constructive.
β’ When crests coincide with troughs, the interference is destructive.
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Co-Propagating Waves
Constructive
Destructive
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An application of destructive interference: getting waves
to cancel each other:
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Adding Multiple Harmonic Waves: Fourier Analysis
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Time and Frequency Descriptions
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Dispersion: when the speed π£π£ ππdepends on the frequency
No dispersion
With dispersion
π£π£ ππ = πππππ π πππ‘π‘πππ π π‘π‘
π£π£ ππ
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Beats:
π¦π¦1 π‘π‘ = π΄π΄ ππππππ ππ1 π‘π‘
π¦π¦2 π‘π‘ = π΄π΄ ππππππ ππ2 π‘π‘
π¦π¦1 π‘π‘ + π¦π¦2 π‘π‘ = 2 π΄π΄ ππππππ12ππ1 β ππ2 π‘π‘ ππππππ
12ππ1 + ππ2 π‘π‘
two co-propagating waves of slightly different frequencies
π‘π‘
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Interference in 2D
π¦π¦1 ππ1 = π΄π΄ ππππππ ππ ππ1
π¦π¦2 ππ2 = π΄π΄ ππππππ ππ ππ2
ππ ππ1 β ππ2 = ππ 2 ππ
ππ ππ1 β ππ2 = ππ 2 ππ + 1
Constructive Interference:
Destructive Interference:
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Total Reflection at an interface
PhET
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Refraction at an interfaceand partial reflection and transmission
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Partial Reflection and Transmission
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Standing Waves: interference between
two counter-propagating waves of the same frequency
π¦π¦1 π₯π₯, π‘π‘ = π΄π΄ ππππππ ππ π₯π₯ β ππ π‘π‘
π¦π¦2 π₯π₯, π‘π‘ = βπ΄π΄ ππππππ ππ π₯π₯ + ππ π‘π‘
π¦π¦1 π‘π‘ + π¦π¦2 π‘π‘ = 2 π΄π΄ πππ π π π ππ π‘π‘ πππ π π π ππ π₯π₯
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ππ πΏπΏ = ππ ππ
π¦π¦1 π‘π‘ + π¦π¦2 π‘π‘ = 2 π΄π΄πππ π π π ππ π‘π‘ πππ π π π ππ π₯π₯
πΏπΏ = ππππ2
ππ = 1, 2, 3, 4, β¦
2 πΏπΏ1 ,
2 πΏπΏ2 ,
2 πΏπΏ3 ,
2 πΏπΏ4 ,ππ =
ππ = 1π£π£
2 πΏπΏ , 2π£π£
2 πΏπΏ , 3π£π£
2 πΏπΏ , 4π£π£
2 πΏπΏ ,
β¦
β¦
fundamental harmonics
π₯π₯ = πΏπΏ πππ π π π ππ πΏπΏ = 0
π₯π₯ = 0 πππ π π π ππ 0 = 0
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PhET
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ππ πΏπΏ = ππππ2
π¦π¦1 π‘π‘ + π¦π¦2 π‘π‘ = 2 π΄π΄ ππππ ππ ππ π‘π‘ ππππππ ππ π₯π₯
ππ πΏπΏ = 2 ππ + 1ππ4
= 2 ππππ4
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Doppler Effect
ππβ² = ππ β π’π’ ππ
= ππ β π’π’πππ£π£
from a moving sourcewith respect to the wave carrier medium,
observer at rest w.r.t to the carrier medium
= ππ 1 βπ’π’π£π£
ππβ² =ππ
1 β π’π’π£π£
ππβ² = ππ 1 βπ’π’π£π£
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Doppler Effectfrom a moving source
with respect to the carrier medium,observer at rest w.r.t to the carrier medium
πππ΄π΄,π΅π΅β² = ππ 1 β
π’π’π£π£
πππ΄π΄,π΅π΅β² =
ππ
1 β π’π’π£π£
π’π’
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Doppler Effect
π₯π₯ππππππππ = βππ + π£π£ π‘π‘
from a moving observerwith respect to the wave carrier medium,source at rest w.r.t to carrier medium
ππβ² = ππ 1 +π’π’π£π£
ππβ² =ππ
1 + π’π’π£π£
π΄π΄π’π’
π₯π₯ππππππ = 0 β π’π’ π‘π‘
βπ’π’ ππβ² = βππ + π£π£ πππ
βπ’π’ ππβ² = βπ£π£ ππ + π£π£ πππ
ππβ² =ππ
1 + π’π’π£π£
π₯π₯ππππππ = π₯π₯ππππππππ
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Velocity of a Wave
π£π£π€π€πππ€π€ππ, πππππππππππ€π€ππππ = π£π£π€π€πππ€π€ππ, ππππππππππππππ
π£π£
π’π’
π£π£π€π€πππ€π€ππ, πππππππππππ€π€ππππ = π£π£ + π’π’
π£π£π’π’
+ π£π£ππππππππππππππ, πππππππππππ€π€ππππ