Lecture 06 wave energy. interference. standing waves.

1. Lecture 6
Wave energy.
Interference. Standing waves.

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. 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. 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. 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. 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. 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. 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. Interference, superposition
Q: What happens when two waves “collide?”
A: They ADD together! We say the waves are “superposed.”
“Constructive”
“Destructive”

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. 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. 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. 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. 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. 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. +x
-x
Standing wave

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. 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. 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. Normal modes 2D
For circular fixed boundary
DEMO:
Normal modes
square surface