Waves 2006 Physics 23. Armen Kocharian Lecture 3: Sep

Transcription

1 Waves 2006 Physics 23 Armen Kocharian Lecture 3: Sep

2 Last Time What is a wave? A "disturbance" that moves through space. Mechanical waves through a medium. Transverse vs. Longitudinal e.g., string vs. sound Sinusoidal each particle in the medium undergoes SHM.

3 Last Time (cont.) Wave function y(x,t) displacement (y) as a function of position along the direction of propagation (x) and time (t). y P x Wave moves with velocity v in +ve x- direction y(x-vt) Do not confuse velocity of wave with velocity of

4 y(x-vt) y Q y v P x Pulse at t=0 y(x,t)=y(x,0) P v Pulse at later time t x vt At time t, an element of the string (P) at some x has the same y position as an element located at x-vt at t=0 (Q). y(x,t)=y(x-vt,0)=y(x-vt)

5 Last time (cont.) Sinusoidal Wave: Wave Equation:

6 Energy Considerations Waves carry energy. Think of a pulse on a string. Energy is transferred from hand to string. Kinetic energy moves down the string.

7 Consider an element of a string in motion (left to right): It moves (accelerates) because each piece of the medium exerts a force on its neighboring piece. What is the force on "a"? What is the power?

8 Power: This is the rate at which work is being done (P=W/t), and the rate at which energy travels down the string. For sinusoidal wave

9 Then and mass length Property of the string Property of the wave This is a general property of mechanical waves: 1. Power proportional to square of amplitude 2. Power proportional to square of frequency

10 Maximum value of power: Average value of power Since average of sin 2 is 1/2

11 Wave reflection When a wave encounters an "obstacle",.i.e., a "change in the medium" something happens. For example: a sound wave hitting a wall is "reflected" a light wave originally traveling in air when it reaches the surface of a lake is partially "reflected" and partially "transmitted".

12 Wave reflection, string Imagine that one end of the string is held fixed:

13 Why is the reflected pulse inverted? Pulse was initially created with upward and then downward force on end of string. When pulse arrives at fixed end, string exerts upward force on support. The end of the string does not move (it is fixed!). By Newton's 3 rd law, support exerts downward force on string. When the top of the pulse arrives, the string exerts a downward force on support. Newton's 3 rd law support exerts upward force on string. Support-to-string force: downward then upward. Opposite order as pulse creation (was upward then downward). Reflected pulse is inverted

14 In this (idealized) situation the reflected wave has the same amplitude (magnitude) and velocity (magnitude) as the incoming wave. No energy is lost in the reflection.

15 Now imagine that one end of the string is free:

16 Why is the reflected pulse not inverted? Pulse was initially created with upward and then downward force on the far end of the string. When pulse first arrives at free end, there is an upward force on the end of the string. When the top of the pulse arrives, the direction of the force becomes downward. upward and then downward force on the free end of the string Forces on free end like at far end where the pulse was 1 st generated. No inversion on reflection

17 Boundary Conditions The properties ("conditions") at the end of the string (or more generally where the medium changes) are called "boundary conditions". This is jargon, but it is used in many places in physics, so try to remember what it means.

18 time Interference Imagine that the incoming pulse is long. Near the boundary at some point we will have a "meeting" of the incoming pulse and the reflected pulse. The deflection of the string will be the sum of the two pulses. (principle of superposition)

19 Principle of Superposition When two (or more) waves overlap, the actual displacement at any point is the sum of the individual displacements. Total displacement First wave Second wave Consequence of the fact that wave equation is linear in the derivatives.

20 Standing Waves Consider a sinusoidal wave traveling to the left: String held fixed at x=0 reflected wave: kx+ωt kx-ωt because reflected wave travels to the right. what about δ? Must choose it to match the boundary conditions!

21 Boundary condition: string is fixed at x=0 Mathematically y(x=0,t) = 0 at all times t But y(x=0,t)=0:

22 2A Nodes X=0 kx=π kx=2π

23 Imagine that string is held at both ends. L=length of the string Nodes at x=0 and x=l Only standing waves of very definite wavelengths (and frequencies) are allowed

24 Normal Modes

25 If you could displace a string in a shape corresponding to one of the normal modes, then the string would vibrate at the frequency of the normal mode Surrounding air would be displaced at the same frequency producing a pure sinusoidal sound wave of the same frequency. In practice when you pluck a guitar string you do not excite a single normal mode. Because you do not displace the string in a perfectly sinusoidal way

26 The displacement of the string can be represented as a sum over the normal modes (Fourier series). adding an infinite numbers of terms you can get the exact shape

27 How to control the frequency of the normal modes Longer strings lower frequencies. Cello vs violin Higher tension (F) higher frequencies. More messive strings lower frequencies.

28 Normal Modes 15.28: T=0.25 s T=0.50 s T=0.75 s T=1.00 s T=1.25 s

29 Normal Modes 15.29: 4.0 s 15.27: a) The wave form for the given times, respectively, is shown. 1.0 ms 2.0 ms 3.0 ms 6.0 s 4.0 ms 5.0 ms 6.0 ms 7.0 ms 1.0 ms 2.0 ms 3.0 ms 4.0 ms 5.0 ms 6.0 ms 7.0 ms

30 Normal Modes a) 15.39: b) Eq. (15.28) gives the general equation for a standing wave on a string: y( xt, ) = ( A sin kx) sinω t SW A SW = 2A, so A= A SW 2 = (5.60 cm) 2 = 2.80 cm c) The sketch in part (a) shows that L = 3( λ 2) k = 2 π λλ, = 2π k Comparison of y( x, t) given in the problem to Eq. (15.28) gives k = rad cm. So, λ = 2π ( rad cm) = cm L = 3(λ 3) = 277 cm

31 Normal Modes d) λ =185 cm, from part (c) ω = 50.0 rad s so f = ω 2π = 7.96Hz period T = 1f= s v = f λ = 1470 cm s e) v y = dy dt = ωa SW sin kx cosω t v y,max = ωa SW = (50.0 rad s)(5.60cm) = 280 cm s f) f 3 = 7.96 Hz = 3f 1,sof 1 = 2.65 Hz f 8 = 8f 1 = 21.2 Hz; ω 8 = 2πf 8 =133 rad s is the fundamental λ = vf= (1470 cm s) (21.2 Hz) = 69.3 cm and k = 2π λ = rad cm. y( xt, ) = (5.60cm) sin ([0.0906rad cm] x) sin ([133rad s] t)

32 Normal Modes 15.42: The condition that x = L is a node becomes kl n = nπ. The wave number and wavelength are related by k n λ n = 2π, and so λ n = 2L n.

33 Normal Modes 15.45: a) λ = 2L= 3.00 m, f = = = 16.0Hz. v (48.0m s) 1 1 2L 2(1.50m) b) λ 3 = λ 1 3 = 1.00 m, f 2 = 3f 1 = 48.0Hz. c) λ 4 = λ 1 4 = 0.75 m, f 3 = 4f 1 = 64.0 Hz.

34 Normal Modes 15.48: a) From comparison with Eq. ( 15.4), A= 0.75 cm, λ = cm = 5.00 cm, f = 125 Hz, T = 1 f = s and v = λf = 6.25 m s. c) To stay with a wave front as t increases, x decreases and so the wave is moving in the d) From Eq. ( ), the tension is F 2 = μ v = x 2 (0.50 kg m) (6.25 m s) = 19.5 N. e) P av = 1 2 μfω 2 A 2 = 54.2 W. - direction.

35 Normal Modes 15.35: The wave equation is a linear equation, as it is linear in the derivatives, and differentiation is a linear operation. Specifically, y x = (y 1 + y 2 ) = y 1 x x + y 2 x. Repeating the differentiation to second order in both x and t, 2 y = 2 y y 2, x 2 x 2 x 2 y1 and 2 2 y t = 2 y y 2 t 2 t. 2 y are given as being solutions to the wave equation; that is, 2 y x 2 = 2 y 1 x y 2 x 2 = = 1 v 2 1 v 2 2 y 1 + t 2 1 v 2 2 y y 2 t 2 t 2 2 y 2 t 2 = 1 v 2 2 y t 2

36 Normal Modes 15.51: a) The functions y( x, t) = A cos ( kx ωt) v y = dy dt = + Aω sin ( kx ωt) v y,max = Aω = 2π fa f = v λ and v = F ( m L), so f = 1 λ v y,max = 2πA FL λ M FL M b) To double v y, max increase F by factor of 4

37 15.62: a) Normal Modes The wave moves in positive x direction with speed v, so we must replace x with x-vt 0 h( L + x vt) y( x, t) = h( L x + vt) 0 for ( x vt) < L L for L < ( x vt) < 0 L for 0 < ( x vt) < L for ( x vt) > L (c) From Eq. (15.21): F(0)(0) = 0 for (x vt) < L y(x, t) y(x, t) F(h L)( hv L) = Fv(h L) 2 for L < (x vt) < 0 P(x, t) = F = x t F( h L)(hv L) = Fv(h L) 2 for 0 < (x vt) < L F(0)(0) = 0 for (x vt) > L d) Thus the instantaneous power is zero except for L < (x vt) < L, where it has the constant value Fv(h L) 2.