Physics 1501 Lecture 28

Transcription

1 Phsics 1501 Lecture 28 Phsics 1501: Lecture 28 Toda s Agenda Homework #10 (due Frida No. 11) Midterm 2: No. 16 Topics 1-D traeling waes Waes on a string Superposition Power Phsics 1501: Lecture 28, Pg 1 Chap. 13: Waes What is a wae? A definition of a wae: A wae is a traeling disturbance that transports energ but not matter. Eamples: Sound waes (air moes back & forth) Stadium waes (people moe up & down) Water waes (water moes up & down) Light waes (what moes??) Animation Phsics 1501: Lecture 28, Pg 2 Page 1

2 Phsics 1501 Lecture 28 Tpes of Waes Transerse: The medium oscillates perpendicular to the direction the wae is moing. Water (more or less) String waes Longitudinal: The medium oscillates in the same direction as the wae is moing Sound Slink Phsics 1501: Lecture 28, Pg 3 Wae Properties Waelength: The distance λ between identical points on the wae. Amplitude: The maimum displacement A of a point on the wae. Amplitude A Waelength λ A Animation Phsics 1501: Lecture 28, Pg 4 Page 2

3 Phsics 1501 Lecture 28 Wae Properties... Period: The time T for a point on the wae to undergo one complete oscillation. Speed: The wae moes one waelength λ in one period T so its speed is = λ / T.! = T Animation Phsics 1501: Lecture 28, Pg 5 = λ / T Wae Properties... We will show that the speed of a wae is a constant that depends onl on the medium, not on amplitude, waelength or period λ and T are related! λ = T or λ = 2π / ω (since T = 2π / ω ) or λ = / f (since T = 1/ f ) Recall f = ccles/sec or reolutions/sec ω = rad/sec = 2πf Phsics 1501: Lecture 28, Pg 6 Page 3

4 Phsics 1501 Lecture 28 Lecture 28, Act 1 Wae Motion The speed of sound in air is a bit oer 300 m/s, and the speed of light in air is about 300,000,000 m/s. Suppose we make a sound wae and a light wae that both hae a waelength of 3 meters. What is the ratio of the frequenc of the light wae to that of the sound wae? (a) About 1,000,000 (b) About.000,001 (c) About 1000 Phsics 1501: Lecture 28, Pg 7 Wae Forms So far we hae eamined continuous waes that go on foreer in each direction! We can also hae pulses caused b a brief disturbance of the medium: And pulse trains which are somewhere in between. Phsics 1501: Lecture 28, Pg 8 Page 4

5 Phsics 1501 Lecture 28 Mathematical Description Suppose we hae some function = f(): f(-a) is just the same shape moed a distance a to the right: 0 Let a=t Then f(-t) will describe the same shape moing to the right with speed. 0 0 =a =t Phsics 1501: Lecture 28, Pg 9 Math... Consider a wae that is harmonic in and has a waelength of λ. A λ If the amplitude is maimum at =0 this has the functional form: 2( $ % ' & # ( ) = Acos! " Now, if this is moing to the right with speed it will be described b: 2) $ % ( & # (,t) = Acos ( ' t)!" Phsics 1501: Lecture 28, Pg 10 Page 5

6 Phsics 1501 Lecture 28 Math... So we see that a simple harmonic wae moing with speed in the direction is described b the equation: 2) $ % ( & # (,t) = Acos ( ' t)!" B using!!" = = T 2# from before, and b defining k! 2" # we can write this as: (,t) = Acos( k "! t) (what about moing in the - direction?) Phsics 1501: Lecture 28, Pg 11 Math Summar Moie (twae) The formula (,t) = Acos( k "! t) describes a harmonic wae of amplitude A moing in the + direction. λ A Each point on the wae oscillates in the direction with simple harmonic motion of angular frequenc ω. The waelength of the wae is "! = 2 k The speed of the wae is =! k The quantit k is often called wae number. Phsics 1501: Lecture 28, Pg 12 Page 6

7 Phsics 1501 Lecture 28 Lecture 28, Act 2 Wae Motion A harmonic wae moing in the positie direction can be described b the equation (,t) = A cos ( k - ωt ) Which of the following equation describes a harmonic wae moing in the negatie direction? (a) (,t) = A sin ( k ωt ) (b) (,t) = A cos ( k + ωt ) (c) (,t) = A cos ( k + ωt ) Phsics 1501: Lecture 28, Pg 13 Traeling 1-D Waes : Traeling Pulse At t = 0 t=t = f() =? (,t) At t = t (moing in + direction) = f(- ) = t = f(-t) 0 = t Phsics 1501: Lecture 28, Pg 14 Page 7

8 Phsics 1501 Lecture 28 Traeling 1-D Waes : Traeling Pulse At t = 0 = f() =? (,t) At t = t (moing in - direction) = f(+ ) = t = f(+t) 0 = t Phsics 1501: Lecture 28, Pg 15 Traeling 1-D Period Wae : Displacement t=0 Waelength λ Amplitude A Period (T) t=t 1 t=t2 t=t 3 (,t) = Acos( k "#t + $ ) k = 2! " Angular frequenc (ω) Frequenc (f) Phase (φ) 2"! = 2" f = T Wae number (k) speed () : = " T = "f = # k Phsics 1501: Lecture 28, Pg 16 Page 8

9 Phsics 1501 Lecture 28 Lecture 28, Act 3 Wae Motion A boat is moored in a fied location, and waes make it moe up and down. If the spacing between wae crests is 20 meters and the speed of the waes is 5 m/s, how long Δt does it take the boat to go from the top of a crest to the bottom of a trough? (a) 2 sec (b) 4 sec (c) 8 sec t t + Δt Phsics 1501: Lecture 28, Pg 17 Waes on a string What determines the speed of a wae? Consider a pulse propagating along a string: Snap a rope to see such a pulse How can ou make it go faster? Animation Phsics 1501: Lecture 28, Pg 18 Page 9

10 Phsics 1501 Lecture 28 Waes on a string... Suppose: The tension in the string is F The mass per unit length of the string is µ (kg/m) The shape of the string at the pulse s maimum is circular and has radius R F µ R Phsics 1501: Lecture 28, Pg 19 Waes on a string... Consider moing along with the pulse Appl F = ma to the small bit of string at the top of the pulse which is moing with Uniform Circular Motion. Phsics 1501: Lecture 28, Pg 20 Page 10

11 Phsics 1501 Lecture 28 Waes on a string... The total force F TOT is the sum of the tension F at each end of the string segment. The total force is in the - direction. F θ θ F F TOT = 2F θ (since θ is small, sin θ ~ θ) Phsics 1501: Lecture 28, Pg 21 Waes on a string... The mass m of the segment is its length (R 2θ) times its mass densit µ. θ m = R 2θ µ 2θ R θ Phsics 1501: Lecture 28, Pg 22 Page 11

12 Phsics 1501 Lecture 28 Waes on a string... The acceleration a of the segment is 2 / R (centripetal) in the - direction. a R Phsics 1501: Lecture 28, Pg 23 Waes on a string... So F TOT = ma becomes: 2 2F" = R2"µ! R F TOT m a 2 F = µ! = F µ tension F mass per unit length µ Phsics 1501: Lecture 28, Pg 24 Page 12

13 Phsics 1501 Lecture 28 Waes on a string... So we find: = F µ Animation tension F mass per unit length µ Making the tension bigger increases the speed. Making the string heaier decreases the speed. As we asserted earlier, this depends onl on the nature of the medium, not on amplitude, frequenc etc of the wae. Phsics 1501: Lecture 28, Pg 25 Lecture 28, Act 4 Wae Motion A hea rope hangs from the ceiling, and a small amplitude transerse wae is started b jiggling the rope at the bottom. As the wae traels up the rope, its speed will: (a) increase (b) decrease (c) sta the same Phsics 1501: Lecture 28, Pg 26 Page 13

14 Phsics 1501 Lecture 28 Superposition Q: What happens when two waes collide? A: The ADD together! We sa the waes are superposed. Phsics 1501: Lecture 28, Pg 27 Aside: Wh superposition works As we will see in the net lecture, the equation goerning waes (a.k.a. the wae equation ) is linear. It has no terms where ariables are squared. For linear equations, if we hae two (or more) separate solutions, f 1 and f 2, then Bf 1 + Cf 2 is also a solution! You hae alread seen this in the case of simple harmonic motion: 2 d 2 = "! 2 linear in! dt = Bsin(ωt)+ Ccos(ωt) Phsics 1501: Lecture 28, Pg 28 Page 14

15 Phsics 1501 Lecture 28 Superposition & Interference We hae seen that when colliding waes combine (add) the result can either be bigger or smaller than the original waes. We sa the waes add constructiel or destructiel depending on the relatie sign of each wae. will add constructiel will add destructiel In general, we will hae both happening Phsics 1501: Lecture 28, Pg 29 Superposition & Interference Consider two harmonic waes A and B meeting. Same frequenc and amplitudes, but phases differ. The displacement ersus time for each is shown below: A(ωt) B(ωt) What does C(t) = A(t) + B(t) look like?? Phsics 1501: Lecture 28, Pg 30 Page 15

16 Phsics 1501 Lecture 28 Superposition & Interference Add the two cures, A = A 0 cos(k ωt) B = A 0 cos (k ωt - φ) Eas, C = A + B C = A 0 (cos(k ωt) + co (k ωt + φ)) Use formula cos(a+b) = 2cos(1/2(a+b))cos(1/2(a-b)) Doing the algebra gies, C = 2 A 0 cos(φ/2) cos(k ωt - φ/2) Phsics 1501: Lecture 28, Pg 31 Superposition & Interference Consider, C = 2 A 0 cos(φ/2) cos(k ωt - φ/2) A(ωt) B(ωt) C(k-ωt) Amp = 2 A 0 cos(φ/2) Phase shift = φ/2 Phsics 1501: Lecture 28, Pg 32 Page 16

17 Phsics 1501 Lecture 28 Lecture 28, Act 5 Superposition You hae two continuous harmonic waes with the same frequenc and amplitude but a phase difference of 170 meet. Which of the following best represents the resultant wae? Original wae (other has different phase) A) B) D) C) E) Phsics 1501: Lecture 28, Pg 33 Page 17