One-Dimensional Wave Propagation (without distortion or attenuation)

Transcription

1 Phsics 306: Waves Lecture 1 1//008 Phsics 306 Spring, 008 Waves and Optics Sllabus To get a good grade: Stud hard Come to class satapal@phsics.gmu.edu Surve of waves One-Dimensional Wave Propagation (without distortion or attenuation) Consider a 1-D disturbance of arbitrar shape propagating along with speed V > 0. We assume there is: 1. no attenuation of this pulse and. no distortion in its shape. (,t) now later (,t) Since the disturbance is moving, it must be a function of position and time. Therefore, we can write (,t) for displacement. the In a moving frame of reference (moving with wavespeed), then there is no time dependence After a time t the pulse has moved along the -ais a distance Vt, where V is the speed of the wave, but in all other respects remains unaltered. Let us introduce a coordinate sstem, S, that travels with the wave at speed V. In this sstem, is no longer of time t and we can write ( ) Page 1 of 1

2 Phsics 306: Waves Lecture 1 1//008 ( ) ( ) No t dependence V S S V P - Vt O O Vt From - Vt we can write, ( ) ( Vt) The general form of a 1-D wavefunction for constant speed is Left moving ( ± Vt) (assuming no distortion). Left moving This means that and t combine to form a single argument ± Vt. (In other words, we onl have to choose a shape that describes the wave profile and then substitute ± Vt for ).) Eamples; all traveling waves: Page of 1

3 Phsics 306: Waves Lecture 1 1//008 sin(k( - Vt)) periodic ( + Vt) ep(-( - Vt) ) But not: a Vt. not a wave traveling to the left or right. How do we find the wave speed, V? Let us write an equation in a factored form to look like, f(const( ± ct)) V c Eample Asin ( k ωt) ω Asin k t k So, V ω/k. Question Let us ask what partial differential equation is satisfied b all such waves, regardless of their particular function (profile)? That is, what partial differential equation has as its solution the function ( ± Vt)? 1 Answer V Proof To see this ( ) ( Vt) ± where ± Vt. Hence, Holding t constant Holding constant ( ) ( ) 1 ± V V Page 3 of 1 (1) ()

4 Phsics 306: Waves Lecture 1 1//008 Taking the nd deriv of (1) Taking the nd deriv of () V ± V ± V V 1, V QED. Eample: Transverse Wave on an Ideal Stretched String µ. We assume the string to have a linear mass densit per unit length, µ mass length and is under a constant tension maintained b equal and oppositel directed forces applied at its ends. If the string is locall displaced sidewas and released ( plucked ), then the disturbance propagates along the string in time. We have the tension, τ, on a string with no friction, etc. displacement, () τ θ 1 d τ +dθ + d Let us appl Newton s Laws to element d of displaced string. We make the following approimations and assumptions: 1. magnitude of τ constant, independent of position, i.e. onl the direction of the vector changes, not the magnitude of the tension, Page 4 of 1

5 Phsics 306: Waves Lecture 1 1//008. θ is small, 3. d moves onl in the transverse direction as a result of the disturbance, 4. motion is onl in the - plane, 5. in addition to the displacement not changing the magnitude of the tension, the linear densit of the string, µ mass/length is fied, and 6. τ is much bigger than F grav, so we ignore gravit. We see that as a result of sidewas displacement, there is an unbalanced force that acts on the string (mass) element d. As a result, there is an acceleration. The -component of force at is T T sinθ Note, the tension T can be large even if θ is small. displacement, () T θ T +dθ+d T() T (+d) T () T(+d) T (+d) T () Since θ is small, d + d sin θ ~ tanθ Hence, T T The net upward force in the -direction is substituting Page 5 of 1

6 Phsics 306: Waves Lecture 1 1//008 df T ( + d) T ( ) T d d T Td Appling Newton s Second Law for the unbalanced force: F ma, df dm a Td µ ( d) t where µ is the mass/length, we ma then write dm µ d since θ is small, as derived from, displacement, () d cosθ θ 1 d We know the Talor epansion for cosine gives 4 θ θ cosθ 1 +! 4! + For small θ: cosθ 1. Hence, d dm µ µ d cosθ Getting back to the result from Newton s nd law, Page 6 of 1

7 Phsics 306: Waves Lecture 1 1//008 T µ So, µ T 1 which looks like: V w Since, V w the medium. T µ, we see that we get the wavespeed from the properties of Harmonic (sinusoidal) Waves Now let us eamine the simplest wave for m: where the profile is a sine or cosine.. These tpes of waves are called harmonic waves. Suppose Asin ( k ωt) where k and ω are constant et to be determined. We can rewrite the equation into our familiar form for traveling waves ω Asin k t k Hence, V w ω +, which means the wave moves to the right. k Fied Time onl. A snapshot in time, reveals the sinusoidal displacement as a function of position Page 7 of 1

8 Phsics 306: Waves Lecture 1 1//008 A λ for complete ccle wavelength 0 π/ π 3π/ π A complete ccle means the argument in the sinusoid has changed b π (for fied time t). Therefore, ( k ( Vt) ) sin( k( ( ± λ) Vt) ) sin( k( Vt) π ) sin ± Therefore, k π k λ π Hence, π λ k Fied Position Now we fi. Let T time for a complete ccle. We can view the wave as fied in position and is periodic in time, i.e. increasing t t + T, the waveform is eactl reproduced. Hence, period We arrive at: ( k ( + V ( t + T ))) sin( k( + Vt) π ) sin + kvt π (3) This is an equation between the period T, the wave velocit V, and the propagation constant k. Let ν # ccles/unit time frequenc, i.e. 1 ν T Page 8 of 1

9 Phsics 306: Waves Lecture 1 1//008 Since V ω (4) k We ma combine with Eq. (3): kvt π with Eq. (4) to get ω k T k π To get ω t wt π Hence, ω π T So, ω πν Since, V w ω, we ma write k V w πν λν π λ Definition of Phase, φ Let ( k ω +φ) Asin t initial phase where φ represents a shift to the left. Let t 0, k integer. Where are the nodes for this harmonic wave, i.e. where/when is there no displacement ( 0 (node))? ( k ω +φ) Asin t 0 Now, the sin() 0 when the argument nπ, where n 0, 1,, Then, Page 9 of 1

10 Phsics 306: Waves Lecture 1 1//008 k ωt + φ nπ So, ( nπ φ) k Eample: Let t 0, k, φ 0.3. Where is 0 (node)? When n 0; -φ/k -0.3/ n 1; ( )/ 1.4 A plot for A 1, ( 0.3) Asin + would be Page 10 of 1

11 Phsics 306: Waves Lecture 1 1//008 k wave number 0 φ 0.3 initial phase A 7 amplitude 70 7 (deg) (rad) Wave Power Consider the wave, the tension vector T 0, and the velocit vector shown below: T 0 θ v The power is delivered from left to right. From Freshman Phsics, we know dr P F v and v dt Hence, dr P F dt dw dt Page 11 of 1

12 Phsics 306: Waves Lecture 1 1//008 where W is the work done. From the diagram, we see If then Also, P F v T v T v 0 sin ( θ ) T0 0 Acos Ak sin cos ( k ωt) ( 90 θ ) Because ( k ωt) is < 0 here. v Aω sin ( k ωt) So, P T A ( Ak sin( k ωt) )( Aω sin( k ωt) ) kωt sin ( k ωt) 0 0 and P avg square of the amplitude 1 kωt A 0 Page 1 of 1