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PHYSICS 149: Lecture 24 • Chapter 11: Waves – 11.8 Reflection and Refraction – 11.10 Standing Waves • Chapter 12: Sound – 12.1 Sound Waves – 12. 4 Standing Sound Waves Lecture 24 Purdue University, Physics 149 1 ILQ 1 A thick string and a thin string made of the same material hliWhihhhldfhave equal tensions. Which has the larger speed of waves on these strings? A) thin string B) thi c k s tr ing C) waves have the same speed Lecture 24 Purdue University, Physics 149 2 ILQ 2 Of these properties of a wave, the one that is independent ofhf the ot hers is its A) frequency B) wavelength C) amplitude D) speed Lecture 24 Purdue University, Physics 149 3 Harmonic Waves y(x,t) = A sin(ωt – kx) A = Amplitude= Maximum displacement of a point on the wave λ = Wavelength: Distance between identical points on the wave T = Period: Time for a point on the wave to undergo one complete oscillation. f = frequency=1/T ω = angular frequency= 2π/T k = wave number = 2π/λ λ ω v = = λ f = T k Lecture 24 Purdue University, Physics 149 4 Wave Speed y(x,t) = Acos(ωt − kx) ωtkx−=constant so ωΔ−tkx Δ =0 gives Δx ω ==v Δtk A wave y = A cos(ωt - kx) travels in +x direction A wave y = A cos(ωt + kx) travels in -x direction Lecture 24 Purdue University, Physics 149 5 Suppperposition Princip le • When two or more waves pass through the same region the actual displacement is the sum of the separate displacements. y′(x,t) = y (x,t) + y (x,t) 1 2 • If two waves pass through the same region they continue to move independently. Lecture 24 Purdue University, Physics 149 6 Interference • Interference describes what happens when two waves are travelling in t he same direct ion t hroug h t he same reg ion o f space. • Su perposition principle: the res ulting displacement is the algebraic sum of their separate displacement yy(x’(x, t) = y1(x, t) + y2(x, t) – It the two waves arrive with the same amplitude (both waves have a crest): constructive interference and the resultant pulse is larger then either pulses. – It the two waves arrive with opposite amplitude (one has a crest, the other a valley): destructive interference the resulting pulse could be zero. Lecture 24 Purdue University, Physics 149 7 Interference and Suppperposition Lecture 24 Purdue University, Physics 149 8 Reflection • When a wave travels from one boundary to another, reflection occurs. Some of the wave travels backwards from the boundary – Traveling from fast to slow inverted – Traveling slow to fast upright Lecture 24 Purdue University, Physics 149 9 Reflection and Transmission z When a wave strikes an obstacle or comes at the end of the medium it is travelling in, it is reflected (at least in part). z If the end of the rope is fixed the reflected pulse is inverted z If the end of the rope is free Node Antinode the reflected pulse is not inv erted Lecture 24 Purdue University, Physics 149 10 Reflected Wave: Inverted or Not? • The reflected wave will be inverted if it reflects from a medium with a higher mass density (that is, with a lower wave speed; recall ). • The reflected wave will not be inverted if it reflects from a medium with a lower mass density (that is, with a higher wave speed). Å This phenomenon can be explained by (1) The principle of superposition at the fixed point at the end, or (2) Newton’s third law fthffor the forces b btetween th thtie string and dthll the wall. Lecture 24 Purdue University, Physics 149 11 ILQ • A transverse wave on a string is described by y(x,t) = A cos(ωt + kx). It arrives at the point x = 0 where the string is fixed in place. Which function describes the reflected wave? a) A cos(ωt + kx) b) A cos(ωt – kx) c) –A cos(ωt – kx) d) –A sin(ωt + kx) Lecture 24 Purdue University, Physics 149 12 Example • You send a wave pulse (upright) down a lightweight rope, which is tied to a very heavy rope. At the boundary, a) the re flected wa ve will be upri gh t an d th e tr an smi tted wave will be inverted. b) the reflected wave will be inverted and the transmitted wave will be upright. c) the reflected wave will be inverted and the transmitted wave will be inverted. d) the reflected wave will be upright and the transmitted wave will be upr ig ht. Lecture 24 Purdue University, Physics 149 13 ILQ • You send a wave pulse (upright) down a very heavy rope, which is tied to a lightweight rope. At the boundary, a) the re flected wa ve will be upri gh t an d th e tr an smi tted wave will be inverted. b) the reflected wave will be inverted and the transmitted wave will be upright. c) the reflected wave will be inverted and the transmitted wave will be inverted. d) the reflected wave will be upright and the transmitted wave will be upr ig ht. Lecture 24 Purdue University, Physics 149 14 Diffraction Diffraction is the spreading of a wave around an obstacle in its path . Diffraction: Bending of waves around obstacles Lecture 24 Purdue University, Physics 149 15 Speed of Waves Depends on Medium When a wave reaches a boundary between two different media the speed and the wavelength change, but the frequency remains the same. v v f = 1 = 2 λ1 λ2 Lecture 24 Purdue University, Physics 149 16 Law of Refraction θ1 θ2 sin(θ ) v 1 = 1 sin(θ2) v2 Angle of incidence and angle of refraction Lecture 24 Purdue University, Physics 149 17 Standing Waves • Standing waves occur when a wave is reflected at a boundary in such a way that the wave appears to stand still. • In a standing wave on a string, every point moves (as a whole) in simple harmonic motion (SHM) with the same frequency. • Every point reaches its maximum amplitude simultaneously, and every point also reaches its minimum amplitude (namely, zero) simultaneously as well. • Nodes are points of zero amplitude (that is, points that never move); antinodes are points of maximum amplitude. The distance between two adjacent nodes is ½ λ. • Fixed end of a string is always a node, because it never moves. Lecture 24 Purdue University, Physics 149 18 Standing Waves • Nodes: points of destructive interference where the cord remains st ill at a ll t imes. We have a lways no des at fixe d ends. • Antinodes: points of constructi ve interference where the cords oscillates with maximum amplitude. • The nodes and antinodes remain in a fixed position for a given frequency. Standing wave can occur at resonant frequencies. • The lowest frequency that produces the standing wave yields a wave with one antinode. If you double the frequency there will be two anti-nodes etc. Lecture 24 Purdue University, Physics 149 19 How Do We Get Standing Waves? • Graphical Understanding Standing wave: superposition of incident and reflected waves • Mathematical Understanding – IidtIncident wave: A cos(ωt + kx) Å tthlftto the left – Reflected wave: –A cos(ωt – kx) Å to the right, inverted – Accordinggpppp, to the principle of superposition, y(x,t) = A cos(ωt + kx) – A cos(ωt – kx) = 2A cos(ωt) sin(kx) Every point moves in SHM! Lecture 24 Purdue University, Physics 149 20 Possible Wavelengths and Frequencies = 2L/1 = 2L/2 = 2L/3 • f1 is called the fundamental = 2L/4 frequency. • f1, 2f1, 3f1,…,nf1,… (tha t is fn ) are called the natural frequencies or resonant frequencies. Don’t forget that • Resonance occurs when a system fixed ends of the string (like a bridge) is driven at one of its are always nodes. natural frequencies. Lecture 24 Purdue University, Physics 149 21 Standing Waves z Shake the end of a string y(,)xt=+ Asin (()ω t kx ) z If the other end is fixed, a wave travels down and it is reflected back ⇒ interference y(,)xt=− A sin(ω t − kx ) nodes z The two waves in ter fere => st andi ng wave yxt(,)= 2 A cos()sin()ω t kx z Nodes when sin (nπ)=0: nnπ λ Anti-nodes x == n = 0120,1,2 k 2 z Anti-nodes sin (nπ)=+/- 1, half way between nodes Lecture 24 Purdue University, Physics 149 22 Standing Waves • The natural frequency is related to the length of the string L. The lowest frequency (first harmonic) has one antinode 1 L = λ1 λ1 = 2L 2 • The second harmonic has two antinodes λ = L 2 • The n-th harmonic 2 λ = L n n Lecture 24 Purdue University, Physics 149 23 Standing Waves z Once you know the wavelength you also know the fddhdiRllfrequency needed to have a standing wave. Recall: v v nv f = f = = = nf n λ 2L 1 λ n z Note that energy is not transmitted by a standing wave. Since the string is at rest at the nodes no energy flows. Lecture 24 Purdue University, Physics 149 24 ILQ • A violin string of length L is fixed at both ends. Which one of these is not a wavelength of a standing wave on the string? a) L b) 2L c) L/2 d) L/3 e) 3L/2 Lecture 24 Purdue University, Physics 149 25 ILQ • A violinist discovers while tuning her violin that her string is flat (has a lower frequency). She should a) tighten the string b) loosen the string c) plftlay faster Lecture 24 Purdue University, Physics 149 26 Standing Waves L = λ / 2 f1 = fundamental frequency (lowest possible) A guitar’s E-stringgg has a length of 65 cm and is stretched to a tension of 82N.
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