2. Physics 151: Lecture 35, Pg 1 Physics 151: Lecture 35 Today’s Agenda l Topics çWaves on a string çSuperposition çPower

3. Physics 151: Lecture 35, Pg 2 Review: Wave Properties... The speed of a wave (v) is a constant and depends only on the medium, not on amplitude (A), wavelength (  or period (T). remember : T = 1/ f and T = 2  /  and T are related ! l Travleing 1-D wave: y(x,t):

4. Physics 151: Lecture 35, Pg 3 Bats can detect small objects such as insects that are of a size on the order of a wavelength. If bats emit a chirp at a frequency of 60 kHz and the speed of soundwaves in air is 330 m/s, what is the smallest size insect they can detect ?  a.1.5 cm  b.5.5 cm  c. 1.5 mm  d.5.5 mm  e.1.5 um  f.5.5 um Example

5. Physics 151: Lecture 35, Pg 4 Write the equation of a wave, traveling along the +x axis with an amplitude of 0.02 m, a frequency of 440 Hz, and a speed of 330 m/sec.  A.y = 0.02 sin [880  (x/330 – t)]  b.y = 0.02 cos [880  x/330 – 440t]  c.y = 0.02 sin [880  (x/330 + t)]  d.y = 0.02 sin [2  (x/330 + 440t)]  e.y = 0.02 cos [2  (x/330 - 440t)] Example

6. Physics 151: Lecture 35, Pg 5 For the transverse wave described by y = 0.15 sin [  (2x - 64 t)/16] (in SI units), determine the maximum transverse speed of the particles of the medium.  a. 0.192 m/s  b. 0.6  m/s  c. 9.6 m/s  d. 4 m/s  e. 2 m/s Example

7. Physics 151: Lecture 35, Pg 6 Lecture 34, Act 4 Wave Motion l A heavy rope hangs from the ceiling, and a small amplitude transverse wave is started by jiggling the rope at the bottom. l As the wave travels up the rope, its speed will: (a) increase (b) decrease (c) stay the same v l Can you calcuate how long will it take for a pulse travels a rope of length L and mass m ?

8. Physics 151: Lecture 35, Pg 7 Superposition l Q: l Q: What happens when two waves “collide” ? l A: l A: They ADD together! çWe say the waves are “superposed”. see Figure 16.8 See text: 16.4 Animation-1 Animation-2

9. Physics 151: Lecture 35, Pg 8 Aside: Why superposition works l It can be shown that the equation governing waves (a.k.a. “the wave equation”) is linear. çIt has no terms where variables are squared. x = Bsin(  t)+ Ccos(  t) l For linear equations, if we have two (or more) separate solutions, f 1 and f 2, then Bf 1 + Cf 2 is also a solution ! l You have already seen this in the case of simple harmonic motion: linear in x !

10. Physics 151: Lecture 35, Pg 9 Superposition & Interference l We have seen that when colliding waves combine (add) the result can either be bigger or smaller than the original waves. l We say the waves add “constructively” or “destructively” depending on the relative sign of each wave. will add constructively will add destructively l In general, we will have both happening see Figure 16.8 See text: 16.4

11. Physics 151: Lecture 35, Pg 10 Superposition & Interference l Consider two harmonic waves A and B meeting. çSame frequency and amplitudes, but phases differ. l The displacement versus time for each is shown below: What does C(t) = A(t) + B(t) look like ?? A(  t) B(  t)

12. Physics 151: Lecture 35, Pg 11 Superposition & Interference l Add the two curves,  A = A 0 cos(kx –  t)  B = A 0 cos (kx –  t -  ) l Easy, çC = A + B  C = A 0 (cos(kx –  t) + cos (kx –  t +  )) çformula cos(a)+cos(b) = 2 cos[ 1/2(a+b)] cos[1/2(a-b)] çDoing the algebra gives,  C = 2 A 0 cos(  /2) cos(kx –  t -  )

13. Physics 151: Lecture 35, Pg 12 Superposition & Interference A(  t) B(  t) l Consider,  C = 2 A 0 cos(  /2) cos(kx –  t -  ) C(kx-  t) Amp = 2 A 0 cos(  /2) Phase shift =  /2

14. Physics 151: Lecture 35, Pg 13 Lecture 35, Act 1 Superposition l You have two continuous harmonic waves with the same frequency and amplitude but a phase difference of 170° meet. Which of the following best represents the resultant wave? A) E) D) C) B) Original wave (other has different phase)

15. Physics 151: Lecture 35, Pg 14 Lecture 35, Act 1 Superposition The equation for adding two waves with different frequencies, C = 2 A 0 cos(  /2) cos(kx –  t -  /2). The wavelength (2  /k) does not change. The amplitude becomes 2A o cos(  /2). With  =170, we have cos(85°) which is very small, but not quite zero. Our choice has same  as original, but small amplitude. D)

16. Physics 151: Lecture 35, Pg 15 Wave Power l A wave propagates because each part of the medium communicates its motion to adjacent parts. çEnergy is transferred since work is done ! l How much energy is moving down the string per unit time. (i.e. how much power ?) P See text: 16.8

17. Physics 151: Lecture 35, Pg 16 Wave Power... l Think about grabbing the left side of the string and pulling it up and down in the y direction. l You are clearly doing work since F. dr > 0 as your hand moves up and down. l This energy must be moving away from your hand (to the right) since the kinetic energy (motion) of the string stays the same. P See text: 16.8

18. Physics 151: Lecture 35, Pg 17 How is the energy moving? l Consider any position x on the string. The string to the left of x does work on the string to the right of x, just as your hand did: x  x F Power P = F. v v see Figure 16-15 See text: 16.8

19. Physics 151: Lecture 35, Pg 18 Power along the string. Since v is along the y axis only, to evaluate Power = F. v we only need to find F y = -Fsin   -F  if  is small. We can easily figure out both the velocity v and the angle  at any point on the string: l If Recall sin   cos   for small   tan   x F v y vyvy  dy dx See text: 16.8

20. Physics 151: Lecture 35, Pg 19 Power... l So: l But last time we showed that and See text: 16.8

21. Physics 151: Lecture 35, Pg 20 Average Power l We just found that the power flowing past location x on the string at time t is given by: l It is generally true that wave power is proportional to the speed of the wave v and its amplitude squared A 2. l We are often just interested in the average power moving down the string. To find this we recall that the average value of the function sin 2 (kx -  t) is 1 / 2 and find that: See text: 16.8

22. Physics 151: Lecture 35, Pg 21 Recap & Useful Formulas: y x A l Waves on a string l General harmonic waves tension mass / length

23. Physics 151: Lecture 35, Pg 22 Lecture 35, Act 2 Wave Power l A wave propagates on a string. If both the amplitude and the wavelength are doubled, by what factor will the average power carried by the wave change ? i.e. P final /P init = X (a) 1/4 (b) 1/2 (c) 1 (d) 2 (e) 4 initial final

24. Physics 151: Lecture 35, Pg 23 3-D Representation Waves, Wavefronts, and Rays l Up to now we have only considered waves in 1-D but we live in a 3-D world. l The 1-D equations are applicable for a 3-D plane wave. l A plane wave travels in the +x direction (for example) and has no dependence on y or z, Wave Fronts RAYS

25. Physics 151: Lecture 35, Pg 24 Waves, Wavefronts, and Rays l Sound radiates away from a source in all directions. l A small source of sound produces a spherical wave. l Note any sound source is small if you are far enough away from it. 3d representation Shading represents density wave fronts rays

26. Physics 151: Lecture 35, Pg 25 Waves, Wavefronts, and Rays l Note that a small portion of a spherical wave front is well represented as a plane wave.

27. Physics 151: Lecture 35, Pg 26 Waves, Wavefronts, and Rays l If the power output of a source is constant, the total power of any wave front is constant. l The Intensity at any point depends on the type of wave.

28. Physics 151: Lecture 35, Pg 27 l You are standing 10 m away from a very loud, small speaker. The noise hurts your ears. In order to reduce the intensity to 1/2 its original value, how far away do you need to stand? Lecture 35, Act 3 Spherical Waves (a) 14 m (b) 20 m (c) 30 m (d) 40 m

29. Physics 151: Lecture 35, Pg 28 Two ropes are spliced together as shown. A short time after the incident pulse shown in the diagram reaches the splice, the ropes appearance will be that in Lecture 35, Act 4 Traveling Waves Can you determine the relative amplitudes of the transmitted and reflected waves ?

30. Physics 151: Lecture 35, Pg 29 l You are standing 0.5 m away from a very large wall hanging speaker. The noise hurts your ears. In order to reduce the intensity you walk back to 1 m away. What is the ratio of the new sound intensity to the original? Lecture 35, Act 3b Plane Waves (a) 1 (b) 1/2 (c) 1/4 (d) 1/8 speaker 1 m

31. Physics 151: Lecture 35, Pg 30 Recap of today’s lecture l Chapter 16 çWaves on a string çSuperposition çPower