(TRAVELLING) 1D WAVES. 1. Transversal & Longitudinal Waves

Transcription

1 (TRAVELLING) 1D WAVES 1. Transversal & Longitudinal Waves

2 Objectives After studying this chapter you should be able to: Derive 1D wave equation for transversal and longitudinal Relate propagation speed to physical parameters Difference between transversal and longitudinal waves Energy flux of a wave

3 Travelling Waves Consider a travelling wave which does not change its shape as its propagate, for ex. A travelling Gaussion wave along the x-axis. Suppose at time t=0, the shape is y = A exp x2 a 2 And suppose, a few moments later (t) it shifts to the right by the amount of b: y = A exp (x b)2 a 2 Thus the shifting (propagation) speed is v= b/t, so we can write (x vt)2 y(x, t) = A exp a 2

4 Travelling Waves Thus, in general a travelling wave along the x-axis with speed v which does not change its shape is given by a function: y=f(x±vt) Question: which of these functions represent a travelling wave to the right (X+) or left (X-): a. y= sin(t)cos(x) b. y=x +2t c. y = x+xt + t 2 d. y = 2(x-t) 2 e. Y = sin (x+t) e. y=2xt+1

5 Travelling Sinusional Waves A special very important travelling waves is a travelling sinusoidal wave: y x, t = A sin(k(x ± vt) + φ 0 ) or y x, t = A cos(k(x ± vt) + φ 0 ) or y x, t = A exp i(k(x ± vt) + φ 0 ) Where φ (x, t) k x ± vt + φ 0 is the phase and φ 0 is the initial phase A period T is the shortest time needed for a point along the wave to return to its original phase ( a full cycle of vibration)

6 Period and Angular Frequency Since the harmonic functions (sin, cos) have 2π period, then at a certain position along the wave: φ x, t + T = φ x, t + 2π Or k x ± v t + T = k x + vt + 2π So kvt = 2π or kv = 2π/T ω Where ω = 2π = 2πf angular frequency with f=1/t T is the frequency. Using this new definition of ω we can also write the travelling wave function as: Y(x,t) = A sin (kx t+ 0 )

7 Wavelength and its Relations At certain time, the closest distance along the wave which has the same phase is called the wavelength ( ) or it may be identified as the spatial period in comparison with the temporal period (T). φ x + λ, t = φ x, t + 2π, or k x + λ ± vt = k x ± vt + 2π, or kλ = 2π or k = 2π λ is a wave number or angular frequency is space! Compare to 2π T A relationship between λ and T: ω = kv 2π T = 2π v λ = vt λ It means as a certain point vibrates a full cycle of vibration, the wave travels a wave length.

8 1D Transversal Wave Equation- The MOdel The rope is under uniform tension The rope is elastic with no stiffness against transveral force The rope is light Only small deviation from its equilibrium position Uniform mass distribution NO en points effects

9 Derivation of 1D transversal Wave Equation Suppose the tension of the rope is T, and the deviation of the rope from its horizontal position is small (in amplitude), thus ds dx and T is constant. T 1 =T(x) y ds T 2 = T(x+dx) x dx x+dx x Equation for the forces on horizontal and vertical directions: F y = T 2 sinθ 2 T 1 sin θ 1 = δm 2 y dt 2 1 F x = T 2 cosθ 2 T 1 cos θ 1 = 0 2 Small deviation so : T 2 cos θ 2 = T 1 cos θ 1 = T 0

10 Derivation of 1D transversal Wave Equation SO in vertical direction: T 0 tanθ 2 T 0 tan θ 1 δm 2 y dt 2 y y T 0 = δm 2 y x x+dx x x dt 2 2 y T 0 dx x 2 = ρdx 2 y dt 2 x Where, δm = ρdx. The (free) wave equation is then : c 2 = T 0 /ρ We will identify the meaning of c! 2 y x y c 2 dt 2 = 0

11 Solution of the Wave Equation We have seen that y=f(x-vt) or y=f(x+vt) is a travelling wave. So it must be the solution of the wave equation : 2 y dx y c 2 t x y = x 2 x 2 y = x t y = t 2 = 0, substitute y = f(x-vt)=f(u) where u=x-vt u f x vt = f u = f u = f (u) u x u f u = u f u u x = f"(u) u f x vt = f u u t = v f u = vf (u) u 2 t 2 y = v t f u = v u f u u t = v2 f"(u), so

12 SOlution To be a solution, then: f u v2 c 2 f u = 0 and 1 v2 c2 = 0 or c = v which explains why the factor in front of t 2 is actualy v 2 or propagation speed. IN case of wave on the rope: v = T 0 ρ strength ofinteraction inertial of medium We can check the same result is obtained when we use y=f(x+vt). Thus the general solution of the wave equation is the linear combination of both: y x, t = f x vt + g(x + vt) Where f and g are functions depending on the local vibtration generating the wave.

13 Propagation direction and speed Function y=f(u), where u=x-vt : phase, will be determined by its phase (u). Suppose a point Q has its phase u=x-vt at t, after a moment later t =t+ t, it will change it phase to u = x -vt =(x+ x) v (t+ t) If we require that this Q must have the same phase as before then u=u -u =0 or u=constant. x-v t =0 or v = x/ t It means that during time t, the position of Q, has shifted by x, with velocity v. Or in general v = dx dt is known as phase velocity of the wave.

14 Propagation direction and speed The direction of propagation can be found by looking at the phase u = x-vt or u=x+vt. The propagation velocity is du = 0, or dx vdt = 0 v = dx direction, or dt dx + vdt = 0 v = dx dt > 0, propagate to the positive < 0, to negative direction.

15 Harmonic Waves The most important form of wave is harmonic waves: y ψ x, t = ψ 0 cos kx ± ωt, or y ψ x, t = ψ 0 sin kx ± ωt, or y ψ x, t = ψ 0 exp i kx ± ωt Where ω = kv = 2π = 2πf, v = λt where T: period T Harmonic waves are the basis functions to express any other functions in Fourier analysis : For periodic functions : Fourier Series For non periodic functions: Fourier Transform

16 Energy Propagation Wave is propagation of a local disturbances The medium does not propagate The energy (and momentum) is propagating Consider a wave travelling to X+ T y Q v y ψ(x, t) x At point Q, the energy current flowing to the right is coming from the tension on the left. The force for vertical vibration at Q is: F y = T sinθ T 0 tan θ = T 0 ψ x x

17 Energy Propagation The rate at which the energy is being transferred at Q (power) is given by: ψ ψ P = F y v y = T 0 x t This is the energy current (power) flowing to the right direction. Since the wave moving to the right is given by ψ(x vt), then we can also rewrite the energy current flowing to the right as: P(x, t) = T 0 v ψ t 2 = T0 v ψ Note: Prove that this last expression is also valid for left-travelling wave! x 2

18 Energy Propagation of Harmonic Waves Suppose the wave is a harmonic wave : ψ = ψ 0 sin(kx ωt) The energy current : P x, t = T 0 v ψ t 2 = T 0 v ω2 ψ m 2 cos 2 (kx ωt) = vt 0 k 2 ψ m 2 cos 2 (kx ωt) = T 0 v ψ x Considering that T 0 = ρv 2, the energy current can also be expressed as: P x, t = T 0 v ψ t 2 = ρvu 2 Where u : vertical vibration speed! Thus the energy current at a point is also related to its vibration speed. 2

19 Average Energy Current In many cases, we are more interested in the average (over time) of the energy current flow: IN case of harmonic waves: < P > = ρv < u 2 > = ρv u2 rms Where rms: root mean squares! u2 rms =< ω 2 ψ 2 0 sin 2 kx ωt > = 1 2 ω2 2 ψ 0 So the average energy current of harmonic waves is : < P > = v ρu2 rms = v < ε > Where < ε > 1 2 ρω2 ψ 0 2. What is this?

20 Average Energy Current energy time = distance energy time distance Thus < > is the average energy density, in this case energy/length of the rope. It means : < ε > = 1 2 ρω2 ψ 2 0 = 1 2 ρu 2 max < > = maksimum kinetic energy density (per unit length!)

21 Energy Current and Energy Density IN vibration we have learnt: Total energy = kinetic energy + potential energy In case, waves on a rope : = max kinetic energy = max potential energy Total energy density = Kinetic Energy Density Max And : = Potential Energy Density Max Average Energy Flow Rate = Propagation Speed * Total Energy Density < P > = v ε Compare to Electrodynamics : J = v ρ (current and charge density)

22 Impedance of Waves As a response to an external force F, the rope vibrates locally and then propagate which creates waves. For resistive medium, the rope response will be linear and characterize with a quantity called : impedance Z as defined : F = ψ Z t = u or F = Zu compared to V = RI in electrity. For a wave travelling to the right: F = T 0 ψ x So : = F vertical

23 Impedance of Waves F Z = ψ t Z = F ψ t = T ψ 0 x ψ t Impedance = Drivingforce "Current" For a wave travelling to the right: ψ = f x vt so that: ψ = f x t x t = v f x ψ and = f x = f, x x x x Z = T 0 v

24 Energy Current, Impedance and Current Using the definition of wave impedance Z, we can rewrite the energy current as: P x, t = T 0 ψ 2 ψ 2 = Z or v t t 2 2 ψ T 0 x P x, t = Z Compared to electricity : P = R I 2 and P = V 2 /R.

25 Longitudinal Waves: in Solid Suppose we have a solid bar under tension, then we propagate a longitudinal wave along the bar. F(x) x F(x+ x) At rest/ equilibrium x Non equilibrium ξ x+ ξ Consider an element with length x and x-section A. Because of the tension F, it moves according to Newton s Law: F x + Δx F x = Δm 2 ξ t 2

26 Wave Equation : Longitudinal Waves on a solid bar Where ξ : displacement from equilibrium position. The mass element : Δm = ρ 0 A Δx If E= the Young modulus of elasticity, then: F Δξ = E A Δx For small vibration, ξ << x. Using this definition, the Newton s law of motion can be written as: EA Δξ Δx x+δx Δξ Δx x Taking the limit x 0: 2 ξ x 2 = 1 2 ξ E/ρ 0 t 2 = ρ 0 AΔx 2 ξ t 2

27 Speed of Waves on Metal Thus the speed of longitudinal waves on a solid is given by : v 2 = E ρ 0 Typical metals have densities about 7000 kg/m 3 and Young modulus E= Pa. So typical sound speed on metals is : v = E = 1011 =4 km/s ρ In reality, the transversal deformation (as given by Poisson s ratio) of the metals must be taken into account for more accurate determination of the speed.

28 Longitudinal Waves: Sound There are some quantities which travel like waves: Displacement of air particles The density of air The pressure of air Air/Gas cannot withstand change in its shape because of external forces. Therefore it cannot propagate a transversal wave. But gas can response to external pressure by changing its volume (or density). The response is characterized by the Bulk Modulus K:

29 Longitudinal Waves: Sound The elastic bulk modulus is given by: K = ρ 0 dp dρ 0 = V 0 dp dv 0 (I) Where sub.-0 refers to equilibrium gas condition, p is the pressure and is the volume density, whereas V: the volume. Since gas is compressed when a pressure is applied then (dp/dv) < 0, so K is always POSITIVE. Before the wave comes in to the gas: equilibrium pressure (p 0 ), density ( 0 ), and p= f( ), f : function. For small displacement small variation of pressure and density: p = p 0 + Δp, ρ = ρ 0 + Δρ where : small variation.

30 Density and Displacement P(x) x P(x+ x) At rest/ equilibrium x Non equilibrium x+ Initial volume : (x,x+ x) A x expanded volume : (x+, x+ x+ (x+ x)) ( x+ (x+ x)- (x))a The mass inside the initial volume : Δm = ρ 0 A Δx The mass inside expanded volume : Δm = ρ A(Δx + (x+ x) (x)) Conservation of mass : ρ 0 = ρ 1 + { (x+ x) (x)}/δx ρ 0 = ρ 1 + ψ x

31 Variation in Pressure and Displacement But ρ = ρ 0 + Δρ, where all variation are small (,, ψ/ x) So ρ 0 = ρ 1 + ψ =(ρ x 0 + Δρ) 1 + ψ x ρ 0 = ρ ψ ψ + Δρ 1 + x x ψ Δρ ρ 0 (II) x It means, when the displacement is large positive then the pressure drops!

32 Equation of Motion The forces on the element of gas: P x A P x + Δx A = Δm 2 ψ t 2 p x Δx A = ρ 0AΔx 2 ψ t 2 p x = ρ 2 ψ 0 t 2 But p = p 0 + Δp, p 0 is constant! So: Δp x = ρ 0 2 ψ t 2 (III)

33 Wave Equation(displacement) From (I): K = ρ 0 dp dρ 0 K ρ 0 Δρ = Δp From (III): Δp x = ρ 0 From (II) : Δρ 2 ψ t 2 ρ 0 ψ x 2 Δρ = ρ 0 x K 2 ψ t 2 2 ψ = ρ 0 x 2 K 2 ψ t 2 or 2 ψ x 2 1 K ρ0 2 ψ t 2 = 0

34 Computing K Speed of sound is then given by:v 2 = K ρ 0 How to compute K for gas? dp K = ρ 0 dρ 0 Newton assumed the compression and rarefaction of gas during the propagation of sound was fast, so he assumed it was an isothermal process. Assuming ideal gas : PV = nrt Where n = M/M 0 where M : mass of the gas, M 0 mass of 1 mole of the gas

35 Speed of sound : Isothermal process So: P = MRT = ρ 0RT, and the Bulk modulus of the gas is given VM 0 M 0 by: K = ρ 0 v = K ρ 0 = RT M 0 p = ρ 0RT, and the sound speed: ρ 0 M 0 Typical air will have M 0 =29 g/mole, in typical room temperature T=25 C = 298 K, and R= 8.31 J/mol K, will give the estimate: v = m/s) RT M 0 = 292 m/s which a bit lower than experiment (340

36 Speed of sound : Adiabatic process Laplace suggest correction to Newton s idea. The propagation speed is fast enough so that heat does not have enough time to flow from the compressed part to the other part. So it is an adiabatic process: PV γ = C=constant Using this adiabatic relationship and the definition of compressibility, it can be shown that in this case: v = γp ρ At standard room temperature using γ = 1.4 ideal gas, P = 1 atm, ρ = 1.3 kg/m 3, gives v = 328 m/s which is better.

37 Energy flux and Impedance We have shown that K ψ Δρ = Δp and : Δρ ρ ρ 0 0 x Thus the variation (excess) pressure Δp is related to the displacement by : Δp = K ψ x The (excess) force is then F = Δp A, the total power P: P = F u = KA ψ ψ (A) x t The (intensity) or power per unit area or energy flux density is then : P ψ ψ = K (B) A x t For left propagating wave eq (A) and (B) change the signs!

38 Energy flux and Impedance Using the form of wave propagating to the right direction ψ = f(x vt), we may rewrite the formula as: P A = Kv ψ x 2 = K v And the characteristic impedance of the wave can be defined as : F/A Z = ψ t For right propagating wave then : Z = K v = γp 0ρ 0 ψ t 2 ψ or Z = K / ψ x t

39 Energy flux and Impedance Using the definition for impedance, we may rewrite the energy flux density as : P A = K v P A ψ t = Kv ψ x 2 = Z ψ t 2 K ψ = x Z 2 (compare to P=ZI in electricity) 2 (compare to P=V 2 /Z in electricity)

40 Summary In general for any 1D waves (longitudinal/transversal) then : v = β ρ Z = β v = βρ Where volume mass density, and β is : a. Transversal waves on a rope : β = T A m 2 b. Transversal waves on a solid : β = G (shear modulus) c. Longitudinal waves on a solid : β = E (Yound modulus) d. Longitudinal waves on air/gas : β = K (bulk modulus) N

41 Summary Energy flux is given by: P A = β v ψ t 2 = β ψ x 2 P A = Z ψ t 2 = 1 Z β ψ x 2

42 Next : Waves at the boundaries Transmission and reflection!