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Lecture 06 wave energy. interference. standing waves.

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- 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

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