lecture 06 wave energy. interference. standing waves
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Lecture 06 wave energy. interference. standing waves.TRANSCRIPT
Lecture 6Wave 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 rope at the bottom. As the wave travels up the rope, its speed will:
(a) increase
(b) decrease
(c) stay the same
v
Fv
Tension is greater near the top because it has to support the weight of rope under it!
is greater near the top.v
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
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.
Power
What is the work done on the red segment by the string to its left?
1 1 tany x
yF F F
x
and is subject to a force whose y component is
F1
F
F1y
θ
1with (tension in the string)xF F
x
y
y
t
The red segment moves in the y direction with velocity
1 1 1 1
Power (how does energy move along the wave)
y y
y yP F v F v P F
x t
Power for harmonic waves
For a harmonic wave,
2 2siny y
P F Fk A kx tx t
, cosy x t A kx t
sin and siny y
kA kx t A kx tx t
Power (energy flow along x direction)
NB: Always positive, as expected
2 2 2, sinP x t F A kx t
kv F
Maximum power where vertical velocity is largest (y = 0) 2sin kx t
cos kx t
Average power for harmonic waves
2 2 2sinP F A kx t Average over time:
2 2
2 0sin 1
sin2 2
d
2 2 2 21 12 2
vP F A A
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.
ACT: Waves and friction
Consider a traveling wave that loses energy to frictionIf it loses half the energy and its shape stays the same, what is amplitude of the wave ?
A. amplitude decreases by ½B. amplitude decreases by 1/√(2)C. amplitude decreases by ¼
Power is proportional the amplitude squared A 2.
Interference, superposition
Q: What happens when two waves “collide?”
A: They ADD together! We say the waves are “superposed.”
“Constructive” “Destructive”
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 nowdisplaced by bothwaves
Superposition of two identical harmonic waves out of phase
Two identical waves out of phase:
1 , cosy x t A kx t 2 , cosy x t A kx t
intermediateconstructive destructive
Wave 2 is little ahead or behind wave 1
Superposition of two identical harmonic waves out of phase: the math
1 , cosy x t A kx t 2 , cosy x t A kx t
1 2, , , cos cos
cos cos
y x t y x t y x t A kx t A kx t
A kx t kx t
cos cos 2cos cos2 2
a b a ba b
, 2 cos cos2 2
y x t A kx t
When interf erence is completely destructive
0 interf erence is completely constructive
Reflected waves: fixed end.
A pulse travels through a rope towards the end that is tied to a hook in the wall (ie, fixed end)
The pulse is inverted (simply because of Newton’s 3rd law!)
Fon wall by string
Fon string by wall
The force by the wall always pulls in the direction opposite to the pulse.
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:
DEMO: Reflection
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
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?
No motion for these points (nodes)
2k
These points oscillate with the maximum possible amplitude (antinodes)
Standing wave
cos cos 2sin sin2 2
a b a ba b
+x
-x
Standing wave
Standing waves and boundary conditions
We obtained
Nodes 0, , ,... 2
x
3Antinodes , ...
4 4x
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.
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 Nodes 0, , ,... 2
x
, ,... f or 1,2,...2 2
L n n
2 f or 1,2,...n
Ln
n
Allowed standing waves (normal modes) between two fixed ends
Mode n = n-th harmonic
DEMO: Normal modes
on string
2 free ends
1 fixed, 1 free
2 free ends
1 fixed, 1 free
2 fixed ends
2 free ends
1 2L 2 L 3
23L
4 2L Normal modes for fixed
ends (lower row)
First harmonic
Second harmonic
…
Normal modes 2DNormal modes 2D
For circular fixed boundary
DEMO: Normal modes square surface