Waves Energy Transfer Interference Reflection and Transmission

Waves Energy Transfer Interference Reflection and Transmission Lana Sheridan De Anza College May 16, 2016

Last time kinds of waves wave speed on a string pulse propagation the wave equation solutions to the wave equation sine waves transverse speed of an element of the medium

Warm Up Questions You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of the oscillations, but do not change anything about the string, what happens to the speed of the wave? (A) it increases (B) it decreases (C) it remains the same

Warm Up Questions You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of the oscillations, but do not change anything about the string, what happens to the wavelength of the wave? (A) it increases (B) it decreases (C) it remains the same

Warm Up Questions If, instead, you increase the tension in the string and keep the frequency of the oscillations constant, what happens to the speed of the wave? (A) it increases (B) it decreases (C) it remains the same

Warm Up Questions If, instead, you increase the tension in the string and keep the frequency of the oscillations constant, what happens to the wavelength of the wave? (A) it increases (B) it decreases (C) it remains the same

Overview energy transfer by a sine wave interference boundary conditions reflection and transmission

Sine waves: Transverse Speed and Transverse Acceleration v y = ωa cos(kx ωt) a y = ω 2 A sin(kx ωt) = ω 2 y If we fix x =const. these are exactly the equations we had for SHM! The maximum transverse speed of a point P on the string is when it passes through its equilibrium position. v y,max = ωa The maximum acceleration occurs when y = A. a y = ω 2 A

Rate of Energy Transfer in Sine Wave Waves do transmit energy. A wave pulse causes the mass at each point of the string to displace from its equilibrium point. At what rate does this transfer happen?

Rate of Energy Transfer in Sine Wave Waves do transmit energy. A wave pulse causes the mass at each point of the string to displace from its equilibrium point. At what rate does this transfer happen? Consider the kinetic and potential energies in a small length of string. Kinetic: Replacing v y : dk = 1 2 (dm)v 2 y dk = 1 2 (dm)aω2 cos 2 (kx ωt)

Rate of Energy Transfer in Sine Wave Potential: du = 1 2 κy 2 where κ is an effective spring constant for the small mass dm. Recall from last lecture on SHM: ω = κ/m, so replace κ = dm ω 2 to get: du = 1 2 (dm)ω2 A 2 sin 2 (kx ωt) Adding du + dk gives de = 1 2 (dm)ω2 A 2

Rate of Energy Transfer in Sine Wave We know the speed of the wave is v = λ T. If we can just find out how much energy is contained in a full wavelength, then we will know the wave power. The mass of string in one wavelength is m = µλ E λ = 1 2 µω2 A 2 λ Power P = E λ T = 1 2 µω2 A 2 λ T Power of a wave on a string: P = 1 2 µω2 A 2 v

Question Quick Quiz 16.5 1 Which of the following, taken by itself, would be most effective in increasing the rate at which energy is transferred by a wave traveling along a string? (A) reducing the linear mass density of the string by one half (B) doubling the wavelength of the wave (C) doubling the tension in the string (D) doubling the amplitude of the wave 1 Serway & Jewett, page 496.

Interference of Waves When two wave disturbances interact with one another they can amplify or cancel out. Waves of the same frequency that are in phase will reinforce, amplitude will increase; waves that are out of phase will cancel out.

Interference of Waves

Interference of Waves Waves that exist at the same time in the same position in space add together. superposition principle If two or more traveling waves are moving through a medium, the resultant value of the wave function at any point is the algebraic sum of the values of the wave functions of the individual waves. This works because the wave equation we are studying is linear. This means solutions to the wave equations can be added: y is the resultant wave function. y(x, t) = y 1 (x, t) + y 2 (x, t)

y 1 the amplitude y2 pulses align, is Interference the of Waves: Interference sum of theconstructive individual amplitudes. When the pulses overlap, the b y 1! y 2 wave function is the sum of the individual wave functions. a When the y 1 crests of the y 2 two pulses align, the amplitude is c sum of the individual the y 1! overlap, y2 When the pulses the amplitudes. b wave function isy 1the sum of! y2 When the pulses nofunctions. longer the individual wave overlap, they have When the crests of not the been two permanently affected by theis pulses align, the amplitude cthe interference. sum of they individual 1! y 2 b amplitudes. y 1! y 2 When the pulses no longer When the crests ofnot the been two overlap, they have d pulses align,ythe amplitude y 1theis permanently 2affected by the sum of the individual cinterference. W pu th bw in wa ath cw pu W th bwa in W th ov pe cw in pu bth ind W dov W pe c pu in

When the crests of the two align, the amplitude is Interference pulses of Waves: Destructive Interference the difference between the bwhen the pulses overlap, the 1! y 2 individual yamplitudes. wave function is the sum y 2 of athe individual wave functions. y1 cwhen the crests of the two pulses align, the is y 1!amplitude y2 the difference between thethe When the pulses overlap, individual amplitudes. bwhen wave function of the pulses no sum longer y 1!isy 2the the individual wavenot functions. overlap, they have been permanently affected by the c interference. When the crests y 1! yof 2 the two bpulses align, the amplitude is y 1! y 2between the the difference When the pulses no longer y 2 have not been amplitudes. overlap, they dindividual y the permanently affected When the crests of the1bytwo interference. pulses align, the amplitude is

Superposition of Sine Waves Consider two sine waves with the same wavelength and amplitude, but different phases, that interfere. y 1 (x, t) = A sin(kx ωt) y 2 (x, t) = A sin(kx ωt + φ) Add them together to find the resultant wave function, using the identity: ( ) ( ) θ ψ θ + ψ sin θ + sin ψ = 2 cos sin 2 2 Then [ ( )] φ y(x, t) = 2A cos 2 New amplitude sin(kx ωt + φ 2 ) Sine oscillation

y y Interference of Two Sine Waves (equal wavelength) b x y y(x, t) = f 180 y [ ( )] φ 2A cos sin(kx ωt + φ 2 2 ) y 1 y 2 c x f 60

Dependence on Phase Difference ( ) The amplitude of the resultant wave is A = 2A cos φ 2, where φ is the phase difference. For what value of φ is A maximized?

Dependence on Phase Difference ( ) The amplitude of the resultant wave is A = 2A cos φ 2, where φ is the phase difference. For what value of φ is A maximized? φ = 0 or φ = 2π, 2π, 4π, etc. The waves are in phase and constructively interfere. position and Standing Waves y y The individual waves are in phase and therefore indistinguishable. a f 0 x Constructive interference: the amplitudes add. y y 1 y 2 y The individual waves are 180 out of phase. b x

Dependence on Phase Difference position and Standing Waves y y The individual waves are in phase and therefore indistinguishable. a If φ = π, π, 3π, 3π, etc. A = 0. x Destructive interference. f 0 Constructive interference: the amplitudes add. b c y y y 1 y 2 y f 180 y y 1 y2 x x The individual waves are 180 out of phase. Destructive interference: the waves cancel. This intermediate result is neither constructive nor destructive. f 60

Phase Differences We can count phase differences in terms of wavelengths also. If two waves have a phase difference of 1 wavelength then φ = 2π. Constructive interference. If two waves have a phase difference of half a wavelength then φ = π. Destructive interference.

Question Here are four possible phase differences between two identical waves, expressed in wavelengths: 0.20, 0.45, 0.60, and 0.80. Rank them according to the amplitude of the resultant wave, greatest first. (A) 0.20, 0.45, 0.60, 0.80 (B) 0.80, 0.60, 0.45, 0.20 (C) (0.20 and 0.80), 0.60, 0.45 (D) 0.45, 0.60, (0.20 and 0.80) 1 Halliday, Resnick, Walker, page 427.

Phasors We can represent this addition 2 with a phasor diagram. Each wave function at point (x, t) is represented by a vector. 16-11 PHASORS 16-11 PHA 42 This projection This matches projection thismatches this displacement displacement of the dot asof the dot as the wave moves the wave through moves it. through it. y y Zero projection, Zero projection, zero displacement zero displacement y y ω ω y 1 y m1 y 1 y m1 x x y 1 = 0 y 1 = 0 ω ω x (a) (b) (b) 2 of sine waves with equal wavelengths

Phasors x y m1 y 1 x ot of the r two waves. 2 y m1 ω Add the vectors to find the sum. Wave 2, delayed by φ radians Wave 1 (d) This is a snapshot of the two phasors for two waves. Wave 2, delayed by φ radians Addin gives result These are the ω y y 2 m2 This is the projections of projection of y φ y' the two phasors. Adding 1 the two ym1 phasors as vectors the resultant Wave 1 gives the resultant phasor of the phasor. (e) resultant wave. (f ) Fig. 16-14 (a) (d) A phasor of magnitude y m1 rotating ω about an origin at angular s sents a sinusoidal wave.the phasor s projection y 1 on the vertical axis represents the d y of a point This through is the which the wave passes. (e) A m2 second phasor, also of angular speed v y nitude y m2 and rotating at a 2 constant y' projection of m angle fφ from the first phasor, represents a second y' phase constant the resultant f.(f) The resultant wave is represented by the vector sum y m of the tw phasor. β y 1 y m1 magnitude y m1 rotating about an origin at angular speed v reprer s projection y 1 on the vertical axis represents the displacement passes. (e) A second phasor, also of angular speed v but of magnt angle f from the first phasor, represents a second wave, with a t wave is represented by the vector sum y m of the two phasors. (f ) In the diagram A = y m is the amplitude of the resulting wave.

Example Two sinusoidal waves y 1 (x, t) and y 2 (x, t) have the same wavelength and travel together in the same direction along a string. Their amplitudes are A 1 = 4.0 mm and A 2 = 3.0 mm, and their phase constants are 0 and π/3 rad, respectively. What are the amplitude A and phase constant φ of the resultant wave? Also give resultant wave function.

Example Two sinusoidal waves y 1 (x, t) and y 2 (x, t) have the same wavelength and travel together in the same direction along a string. Their amplitudes are A 1 = 4.0 mm and A 2 = 3.0 mm, and their phase constants are 0 and π/3 rad, respectively. What are the amplitude A and phase constant φ of the resultant wave? Also give resultant wave function. A = 6.1 mm ; φ = 0.44 rad

Example Two sinusoidal waves y 1 (x, t) and y 2 (x, t) have the same wavelength and travel together in the same direction along a string. Their amplitudes are A 1 = 4.0 mm and A 2 = 3.0 mm, and their phase constants are 0 and π/3 rad, respectively. What are the amplitude A and phase constant φ of the resultant wave? Also give resultant wave function. A = 6.1 mm ; φ = 0.44 rad y(x, t) = (6.1 mm) sin(kx ωt + 0.44)

Wave Reflection

Boundaries and Wave Reflection and Transmission When waves reach the end of their medium, or move from one medium to another, they can be reflected. The behavior is different in difference circumstances. We can describe the different circumstances mathematically using boundary conditions on our wave function. These will help us to correctly predict how a wave will reflect or be transmitted.

Wave Reflection from a fixed end point a b c Incident pulse Reflected pulse The reflected pulse Figure is inverted. 16.13 How The does reflection this happen? of a traveling pulse at the fixed 16.4 Re The travelin without inte wave is affec a pulse trave Figure 16.13 occurs: the pulse moves Notice th follows. Whe an upward f equal-magn This downw Now cons that is free

Wave Reflection from a fixed end point The boundary condition for a fixed end point at position x = 0 is: y(x = 0, t) = 0 At any time, the point of the string at x = 0 cannot have any vertical displacement. It is tied to a wall! The wave function for single pulse on the string does not satisfy this boundary condition. y 1 (x, t) = f (x vt) This pulse will continue in the +x direction forever, past the end of the string. Makes no sense.

Wave Reflection from a fixed end point The boundary condition for a fixed end point at position x = 0 is: y(x = 0, t) = 0 At any time, the point of the string at x = 0 cannot have any vertical displacement. It is tied to a wall! The wave function for single pulse on the string does not satisfy this boundary condition. y 1 (x, t) = f (x vt) This pulse will continue in the +x direction forever, past the end of the string. Makes no sense. What if we imagine the string continues inside the wall, and there is a pulse traveling behind the wall in the x direction?

1 Wall at x = 2.5. Digrams by Michal Fowler http://galileo.phys.virginia.edu

Wave Reflection from a fixed end point If we allow another wave function: y 2 (x, t) = f ( x vt) the total wave function will satisfy the boundary condition! y(x, t) = y 1 (x, t) + y 2 (x, t) y(x, t) = f (x vt) + [ f ( x vt)] y(x = 0, t) = 0 However, f ( x vt) corresponds to an inverted wave pulse. The reflected pulse is inverted.

Wave Reflection from a fixed end point The reflected pulse is inverted. a b c Incident pulse Reflected pulse Figure 16.13 The reflection 16.4 Re The travelin without inte wave is affec a pulse trave Figure 16.13 occurs: the pulse moves Notice th follows. Whe an upward f equal-magn This downw Now cons

reflected pulse is inverted, but its Wave Reflection shape from is otherwise a freely unchanged. movable end point In this case, reflected pulse is not inverted. a b c Incident pulse Reflected pulse Figure 16.14 The reflection of a traveling pulse at the free end of vertically on a sm time it is not inve end of the string the incoming pu the ring back do not inverted and Finally, consid two extremes. In part undergoes t ary. For instance 16.15. When a pu two strings, part the heavier strin earlier in the cas The reflected 16.5, we show th ing to the princ reflected pulse a

Wave Reflection from a freely movable end point Now we have a different boundary condition. The slope of the string at the boundary must be zero. y x = 0 x=0 This ensures that the string will stay attached to the wall and there will not be an infinite force on the last tiny bit of string. To satisfy this boundary condition, imagine there is another pulse that is upright but moving in the x direction.

Wave Reflection from a freely movable end point Imagine the free end of the string at x = 2.5. The slope there is zero at all times.

Wave Reflection from a freely movable end point The new boundary condition is satisfied if y 2 = f ( x vt): y(x, t) = f (x vt) + f ( x vt) y(x, t) x = f (x vt) x y(x, t) f (x vt) = x x y x = 0 x=0 f ( x vt) + x ( + f (x vt) x ) The pulse f ( x vt) is not inverted.

Transmitted and Reflected Waves at a Boundary If two ropes of different linear mass densities, µ 1 and µ 2 are attached together, an incoming pulse will be partially transmitted and partially reflected. The boundary conditions here are different again. Now the slope of the string at the boundary should be zero and the displacements to at the boundary must be the same (otherwise the string breaks).

Transmitted and Reflected Waves at a Boundary From those boundary conditions it is possible to deduce the behavior: µ 1 < µ 2 µ 1 > µ 2 16.5 Rate of Energy Transfer by Sinusoidal Waves on Strings Incident pulse Incident pulse The reflected pulse is inverted and a non-inverted transmitted pulse moves on the heavier string. a The reflected pulse is not inverted and a transmitted pulse moves on the lighter string. a b Figure 16.15 (a) A pulse traveling to the right on a light string approaches the junction with a heavier string. (b) The situation after the pulse reaches the junction. b Figure 16.16 (a) A pulse traveling to the right on a heavy string approaches the junction with a lighter string. (b) The situation after the pulse reaches the junction. According 1 Serway to & Equation Jewett, 16.18, pagethe 495. speed of a wave on a string increases as the

Summary energy transfer by a sine wave interference reflection and transmission Homework Serway & Jewett: Ch 16, onward from page 499. Probs: 33, 35, (if not done already 61) Ch 18, onward from page 555. OQs: 9; CQs: 9; Probs: 1, 3, 7, 9, 11