Chapter 16 - Waves. I m surfing the giant life wave. -William Shatner. David J. Starling Penn State Hazleton PHYS 213. Chapter 16 - Waves

Transcription

1 I m surfing the giant life wave. -William Shatner David J. Starling Penn State Hazleton PHYS 213

2 There are three main types of waves in physics: (a) Mechanical waves: described by Newton s laws and propagate through matter, such as water, sound and seismic waves.

3 There are three main types of waves in physics: (a) Mechanical waves: described by Newton s laws and propagate through matter, such as water, sound and seismic waves. (b) Electromagnetic waves: described by Maxwell s equations and propagate through vacuum at the speed of light, such as x-rays, gamma rays, radio waves and visible light.

4 There are three main types of waves in physics: (a) Mechanical waves: described by Newton s laws and propagate through matter, such as water, sound and seismic waves. (b) Electromagnetic waves: described by Maxwell s equations and propagate through vacuum at the speed of light, such as x-rays, gamma rays, radio waves and visible light. (c) Matter waves: described my quantum mechanics, these waves explain the wave nature of fundamental particles (electrons, protons, etc).

5 Mechanical waves come in two flavors: transverse

6 Mechanical waves come in two flavors: transverse longitudinal

7 Mechanical waves come in two flavors: transverse longitudinal The oscillations of matter are perpendicular or parrallel to the motion of the wave.

8 All waves satisfy the so-called wave equation. 2 y x 2 = 1 2 y v 2 t 2

9 All waves satisfy the so-called wave equation. 2 y x 2 = 1 2 y v 2 t 2 y is the transverse or longitudinal displacement x is the direction of travel v, a constant, is the speed of the wave

10 All waves satisfy the so-called wave equation. 2 y x 2 = 1 2 y v 2 t 2 y is the transverse or longitudinal displacement x is the direction of travel v, a constant, is the speed of the wave We are looking for y(x, t).

11 Mechanical waves are described by a traveling sine wave. y(x, t) = y m sin(kx ωt)

12 Mechanical waves are described by a traveling sine wave. y(x, t) = y m sin(kx ωt) y m : amplitude (maximum displacement) k = 2π/λ: wavenumber λ: wavelength ω = 2πf : angular frequency

13 Reminder: frequency, period and angular frequency are all related: ω = 2πf f = 1 T ω = 2π T

14 Reminder: frequency, period and angular frequency are all related: ω = 2πf f = 1 T ω = 2π T The wave number k = 2π/λ is a spatial frequency : y(x, t = 0) = y m sin(kx + 0).

15 Each part of the rope is confined to its x position and travels up and down like a harmonic oscillator: y(x = 0, t) = y m sin( ωt)

16 Each part of the rope is confined to its x position and travels up and down like a harmonic oscillator: y(x = 0, t) = y m sin( ωt) The speed of this part of the rope is just u = dy(t) dt = ωy m cos(ωt)

17 We can find the speed of the wave from the argument of the sine function sin(kx ωt), known as the phase of the wave.

18 We can find the speed of the wave from the argument of the sine function sin(kx ωt), known as the phase of the wave. kx ωt = constant d (kx ωt) dt = 0

19 We can find the speed of the wave from the argument of the sine function sin(kx ωt), known as the phase of the wave. kx ωt = constant d (kx ωt) dt = 0 k dx dt ω = 0 dx dt = v = ω k = λ T = λf

20 We can derive the wave speed for a stretched rope with tension τ and mass density µ kg/m.

21 We can derive the wave speed for a stretched rope with tension τ and mass density µ kg/m. F = ma

22 We can derive the wave speed for a stretched rope with tension τ and mass density µ kg/m. F = ma 2τ sin(θ) = m v2 R

23 We can derive the wave speed for a stretched rope with tension τ and mass density µ kg/m. F = ma 2τ sin(θ) = m v2 R µ l v2 τ(2θ) R

24 We can derive the wave speed for a stretched rope with tension τ and mass density µ kg/m. F = ma 2τ sin(θ) = m v2 R µ l v2 τ(2θ) R τ l µ l v2 R R

25 We can derive the wave speed for a stretched rope with tension τ and mass density µ kg/m. F = ma 2τ sin(θ) = m v2 R µ l v2 τ(2θ) R τ l µ l v2 R R v = τ/µ

26 Lecture Question 16.1 Alice and Bob are floating on a quiet river. At one point, they are 5.0 m apart when a speed boat passes. After the boat passes, they begin bobbing up and down at a frequency of 0.25 Hz. Just as Alice reaches her highest level, Bob is at his lowest level. As it happens, they are always within one wavelength. What is the speed of these waves? (a) 1.3 m/s (b) 2.5 m/s (c) 3.8 m/s (d) 5.0 m/s (e) 7.5 m/s

27 Waves transfer energy in the direction of travel. The rate of energy transfer is the power.

28 Waves transfer energy in the direction of travel. The rate of energy transfer is the power. The mass in region b has K but no U. The mass in region a has U but no K.

29 P avg = ( ) dk + dt avg ( du dt ) avg

30 ( ) dk P avg = dt ( ) dk = 2 dt avg + avg ( ) du dt avg

31 ( ) dk P avg = dt ( ) dk = 2 dt = 2 ( 1 2 avg + avg (µ dx) u2 dt ( ) du dt avg ) avg

32 ( ) dk P avg = dt ( ) dk = 2 dt = 2 ( 1 2 avg + avg (µ dx) u2 dt ( ) du dt avg ) avg = ( µv[ ωy m cos(kx ωt)] 2) avg

33 ( ) dk P avg = dt ( ) dk = 2 dt = 2 ( 1 2 avg + avg (µ dx) u2 dt ( ) du dt avg ) avg = ( µv[ ωy m cos(kx ωt)] 2) avg = µvω 2 y 2 m ( cos 2 (kx ωt) ) avg

34 ( ) dk P avg = dt ( ) dk = 2 dt = 2 ( 1 2 avg + avg (µ dx) u2 dt ( ) du dt avg ) avg = ( µv[ ωy m cos(kx ωt)] 2) avg = µvω 2 y 2 m ( cos 2 (kx ωt) ) avg P avg = 1 2 µvω2 y 2 m

35 ( ) dk P avg = dt ( ) dk = 2 dt = 2 ( 1 2 avg + avg (µ dx) u2 dt ( ) du dt avg ) avg = ( µv[ ωy m cos(kx ωt)] 2) avg = µvω 2 y 2 m ( cos 2 (kx ωt) ) avg P avg = 1 2 µvω2 y 2 m The wave generates power in proportion to its mass, velocity and the square of the frequency and amplitude.

36 When two waves on a string overlap, their displacements add algebraically resulting in interference: y (x, t) = y 1 (x, t) + y 2 (x, t)

37 When two waves on a string overlap, their displacements add algebraically resulting in interference: y (x, t) = y 1 (x, t) + y 2 (x, t) Note: the waves emerge without alteration.

38 Special case: two waves of equal magnitude, wavelength and velocity travel along the same string. y (x, t) = y 1 (x, t) + y 2 (x, t) = y m sin(kx ωt) + y m sin(kx ωt + φ)

39 Special case: two waves of equal magnitude, wavelength and velocity travel along the same string. y (x, t) = y 1 (x, t) + y 2 (x, t) = y m sin(kx ωt) + y m sin(kx ωt + φ) = y m [sin(kx ωt + φ/2 φ/2) + sin(kx ωt + φ/2 + φ/2)]

40 Special case: two waves of equal magnitude, wavelength and velocity travel along the same string. y (x, t) = y 1 (x, t) + y 2 (x, t) = y m sin(kx ωt) + y m sin(kx ωt + φ) = y m [sin(kx ωt + φ/2 φ/2) + sin(kx ωt + φ/2 + φ/2)]

41 Special case: two waves of equal magnitude, wavelength and velocity travel along the same string. y (x, t) = y 1 (x, t) + y 2 (x, t) = y m sin(kx ωt) + y m sin(kx ωt + φ) = y m [sin(kx ωt + φ/2 φ/2) + sin(kx ωt + φ/2 + φ/2)] = [2y m cos(φ/2)] sin(kx ωt + φ/2) }{{}}{{} amplitude oscillation

42 y (x, t) = [2y m cos(φ/2)] sin(kx ωt + φ/2)

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44 We can represent a wave with a phasor, a vector of length y m that rotates about the origin at a frequency of ω.

45 We can represent a wave with a phasor, a vector of length y m that rotates about the origin at a frequency of ω. The vertical projection is the displacement of the wave at a particular point.

46 For two waves, you add the phasors to get the resulting displacement.

47 Chapter 16 Lecture Question 16.2 Two waves are traveling along a string. The left wave is traveling to the right at 0.5 cm/s and the right wave is traveling to the left at 2.0 cm/s. At what elapsed time will the two waves completely overlap and what will the maximum amplitude be at that time? (a) 2.0 s, 1.5 cm (b) 1.6 s, 2.5 cm (c) 1.0 s, 1.5 cm (d) 1.0 s, 2.5 cm (e) 1.3 s, 0.0 cm

48 When two waves travel on the same string in opposite directions, the result is a standing wave.

49 When two waves travel on the same string in opposite directions, the result is a standing wave. Nodes are spots where the displacement is always zero, and anti-nodes are the spots of maximum amplitude.

50 Algebraically, a standing wave looks like: y (x, t) = y 1 (x, t) + y 2 (x, t) = y m sin(kx ωt) + y m sin(kx + ωt)

51 Algebraically, a standing wave looks like: y (x, t) = y 1 (x, t) + y 2 (x, t) = y m sin(kx ωt) + y m sin(kx + ωt) = y m [sin(kx) cos(ωt) cos(kx) sin(ωt) + sin(kx) cos(ωt) + cos(kx) sin(ωt)]

52 Algebraically, a standing wave looks like: y (x, t) = y 1 (x, t) + y 2 (x, t) = y m sin(kx ωt) + y m sin(kx + ωt) = y m [sin(kx) cos(ωt) cos(kx) sin(ωt) + sin(kx) cos(ωt) + cos(kx) sin(ωt)] = 2y m sin(kx) cos(ωt) }{{}}{{} amplitude oscillation

53 Algebraically, a standing wave looks like: y (x, t) = y 1 (x, t) + y 2 (x, t) = y m sin(kx ωt) + y m sin(kx + ωt) = y m [sin(kx) cos(ωt) cos(kx) sin(ωt) + sin(kx) cos(ωt) + cos(kx) sin(ωt)] = 2y m sin(kx) cos(ωt) }{{}}{{} amplitude oscillation The amplitude term gives the nodes and anti-nodes.

54 A node is when the amplitude is always zero: sin(kx) = 0 kx = nπ for n = 0, 1, 2,... x = nπ/k = nλ/2

55 A node is when the amplitude is always zero: sin(kx) = 0 kx = nπ for n = 0, 1, 2,... x = nπ/k = nλ/2 An anti-node is when the amplitude is maximum: sin(kx) = 1 kx = (n + 1/2)π for n = 0, 1, 2,... x = (n + 1/2)π/k = (n + 1/2)λ/2

56 When a rope reflects at a barrier, the result depends on the nature of the barrier.

57 When a rope oscillates with two fixed points, certain frequencies called harmonics result in standing waves with nodes and large anti-nodes.

58 When a rope oscillates with two fixed points, certain frequencies called harmonics result in standing waves with nodes and large anti-nodes. λ = 2L n for n = 1, 2, 3... f = v λ = n v 2L

59 When a rope oscillates with two fixed points, certain frequencies called harmonics result in standing waves with nodes and large anti-nodes. λ = 2L n for n = 1, 2, 3... f = v λ = n v 2L If we know the wave velocity (string tension and density), we can predict the harmonic frequencies.

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