Lecture 6
Wave energy.
Interference. Standing waves.
ACT: Wave Motion
A heavy rope hangs from the ceiling, and a small
amplitude transverse wave is started by jiggling
the rop...
Wave energy
• Work is clearly being done: F.dr > 0 as hand
moves up and down.
• This energy must be moving away from your
...
Transfer of energy
The string to the left of x does work on the string to
the right of x, just as your hand did:

x

Energ...
Power
y

x

F

θ
F1

F1y

What is the work done on the red segment by the string to its left?
The red segment moves in the...
Power for harmonic waves
For a harmonic wave, y ( x ,t ) = A cos ( kx − ωt )
∂y
= −kA sin ( kx − ωt )
∂x

P = −F

and

∂y
...
Average power for harmonic waves
Average over time:

P = µF ω2A 2 sin2 ( kx − ωt )
2π

sin α
2

P =

1
1v 2 2
µF ω2A 2 =
ω...
ACT: Waves and friction
Consider a traveling wave that loses energy to friction
If it loses half the energy and its shape ...
Interference, superposition
Q: What happens when two waves “collide?”
A: They ADD together! We say the waves are “superpos...
Aside: Why superposition works
The wave equation is linear: It has no terms where variables
are squared.
If f1 and f2 are ...
Superposition of two identical harmonic waves
out of phase
Two identical waves out of phase:
y1 ( x ,t ) = A cos ( kx − ωt...
Superposition of two identical harmonic waves
out of phase: the math
y1 ( x ,t ) = A cos ( kx − ωt )

y2 ( x ,t ) = A cos ...
Reflected waves: fixed end.

DEMO:
Reflection

A pulse travels through a rope towards the end
that is tied to a hook in th...
Reflected waves: free end.
A pulse travels through a rope towards the end that is tied to a ring that
can slide up and dow...
Standing waves
A wave traveling along the +x direction is reflected at a fixed point. What is
the result of the its superp...
+x

-x

Standing wave
Standing waves and boundary conditions
We obtained y ( x ,t ) = 2A sin ( kx ) sin ( ωt )

Nodes
Antinodes

λ
x = 0, , λ,.....
DEMO:
Normal modes
on string

Normal modes

Which standing waves can I have for a string of length L fixed
at both ends?
I...
1 fixed, 1
free

2 free
ends
2 fixed
ends

λ1 = 2L

λ2 = L

First
harmonic

Second
harmonic…

λ3 =

2L
3

λ4 =

L
2

Norma...
Normal modes 2D

For circular fixed boundary

DEMO:
Normal modes
square surface
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Lecture 06 wave energy. interference. standing waves.

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Lecture 06 wave energy. interference. standing waves.

  1. 1. Lecture 6 Wave energy. Interference. Standing waves.
  2. 2. ACT: Wave Motion A heavy rope hangs from the ceiling, and a small amplitude transverse wave is started by jiggling the rope at the bottom. As the wave travels up the rope, its speed will: (a) increase (b) decrease (c) stay the same Tension is greater near the top because it has to support the weight of rope under it! v = F µ ⇒ v is greater near the top. v
  3. 3. Wave energy • Work is clearly being done: F.dr > 0 as hand moves up and down. • This energy must be moving away from your hand (to the right) since the kinetic energy (motion) of the end of the string grabbed by the hand stays the same. P
  4. 4. Transfer of energy The string to the left of x does work on the string to the right of x, just as your hand did: x Energy is transferred or propagated.
  5. 5. Power y x F θ F1 F1y What is the work done on the red segment by the string to its left? The red segment moves in the y direction with velocity ∂y ∂t and is subject to a force whose y component is F1 y = − F1x tan θ = −F with F1x = F Power (how does energy move along the wave) r r ∂y ∂y P = F1 × 1 = F1 yv1 y v P = −F ∂x ∂t ∂y ∂x (tension in the string)
  6. 6. Power for harmonic waves For a harmonic wave, y ( x ,t ) = A cos ( kx − ωt ) ∂y = −kA sin ( kx − ωt ) ∂x P = −F and ∂y = ωA sin ( kx − ωt ) ∂t ∂y ∂y = Fk ωA 2 sin2 ( kx − ωt ) ∂x ∂t ω ω k = = v P ( x ,t ) = µF ω2A 2 sin2 ( kx − ωt ) F µ cos ( kx − ωt ) sin2 ( kx − ωt ) Power (energy flow along x direction) NB: Always positive, as expected Maximum power where vertical velocity is largest (y = 0)
  7. 7. Average power for harmonic waves Average over time: P = µF ω2A 2 sin2 ( kx − ωt ) 2π sin α 2 P = 1 1v 2 2 µF ω2A 2 = ωA 2 2µ ∫ = 0 sin2 αd α 2π = Average power for harmonic waves It is generally true (for other wave shapes) that wave power is proportional to the speed of the wave v and its amplitude squared A2. 1 2
  8. 8. ACT: Waves and friction Consider a traveling wave that loses energy to friction If it loses half the energy and its shape stays the same, what is amplitude of the wave ? A. amplitude decreases by ½ B. amplitude decreases by 1/√(2) C. amplitude decreases by ¼ Power is proportional the amplitude squared A 2.
  9. 9. Interference, superposition Q: What happens when two waves “collide?” A: They ADD together! We say the waves are “superposed.” “Constructive” “Destructive”
  10. 10. Aside: Why superposition works The wave equation is linear: It has no terms where variables are squared. If f1 and f2 are solution, then Bf1 + Cf2 is also a solution! These points are now displaced by both waves
  11. 11. Superposition of two identical harmonic waves out of phase Two identical waves out of phase: y1 ( x ,t ) = A cos ( kx − ωt ) constructive y2 ( x ,t ) = A cos ( kx − ωt + φ ) destructive intermediate Wave 2 is little ahead or behind wave 1
  12. 12. Superposition of two identical harmonic waves out of phase: the math y1 ( x ,t ) = A cos ( kx − ωt ) y2 ( x ,t ) = A cos ( kx − ωt + φ ) y ( x ,t ) = y1 ( x ,t ) + y2 ( x ,t ) = A cos ( kx − ωt ) + A cos ( kx − ωt + φ ) = A cos ( kx − ωt ) + cos ( kx − ωt + φ )     a +b cos ( a ) + cos ( b ) = 2cos   2   a −b  ÷cos  ÷   2  φ   φ y ( x ,t ) = 2A cos  ÷cos  kx − ωt + ÷ 2 2  When φ = π interference is completely destructive φ =0 interference is completely constructive
  13. 13. Reflected waves: fixed end. DEMO: Reflection A pulse travels through a rope towards the end that is tied to a hook in the wall (ie, fixed end) Fon wall by string Fon string by wall The force by the wall always pulls in the direction opposite to the pulse. The pulse is inverted (simply because of Newton’s 3rd law!) Another way (more mathematical): Consider one wave going into the wall and another coming out of the wall. The superposition must give 0 at the wall. Virtual wave must be inverted:
  14. 14. Reflected waves: free end. A pulse travels through a rope towards the end that is tied to a ring that can slide up and down without friction along a vertical pole (ie, free end) No force exerted on the free end, it just keeps going Fixed boundary condition Free boundary condition
  15. 15. Standing waves A wave traveling along the +x direction is reflected at a fixed point. What is the result of the its superposition with the reflected wave? y1 ( x ,t ) = A cos ( kx − ωt ) y2 ( x ,t ) = −A cos ( kx + ωt ) y ( x ,t ) = A cos ( kx − ωt ) − cos ( kx + ωt )    y ( x ,t ) = 2A sin ( kx ) sin ( ωt ) k = 2π λ Standing wave λ 3λ If kx = 0, π ,2π ,... ⇔ x = 0, , λ, ... 2 2 If kx = π 3π λ 3λ , ,... ⇔ x = , ... 2 2 4 4  a +b   a −b  cos ( a ) − cos ( b ) = 2 sin  ÷sin  ÷  2   2  y ( x ,t ) = 0 No motion for these points (nodes) y ( x ,t ) = ±2A cos ( ωt ) These points oscillate with the maximum possible amplitude (antinodes)
  16. 16. +x -x Standing wave
  17. 17. Standing waves and boundary conditions We obtained y ( x ,t ) = 2A sin ( kx ) sin ( ωt ) Nodes Antinodes λ x = 0, , λ,... 2 x = λ 3λ , ... 4 4 We need fixed ends to be nodes and free ends to be antinodes! Big restriction on the waves that can “survive” with a given set of boundary conditions.
  18. 18. DEMO: Normal modes on string Normal modes Which standing waves can I have for a string of length L fixed at both ends? I need nodes at x = 0 and x = L L= λ λ , λ,... = n 2 2 λn = 2L n Nodes λ x = 0, , λ,... 2 for n = 1,2,... for n = 1,2,... Allowed standing waves (normal modes) between two fixed ends Mode n = n-th harmonic
  19. 19. 1 fixed, 1 free 2 free ends 2 fixed ends λ1 = 2L λ2 = L First harmonic Second harmonic… λ3 = 2L 3 λ4 = L 2 Normal modes for fixed ends (lower row)
  20. 20. Normal modes 2D For circular fixed boundary DEMO: Normal modes square surface

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