General Physics 3: Physics 207 Waves, Optics and Modern Physics Dr. Cerne (pronounced chair-nay) jcerne@buffalo.edu Office Hours: 128 Fronczak Hall How about: Mon 2-3 pm? Wed 3-4:pm? No Recitations this week. Start next week. Physics 207 Homepage!!!: http://www.physics.buffalo.edu/phy207 Recipe for Success in Physics 207 1. Prepare for lectures • attempt assigned reading before class • be active in class (paying attention to lecture is more important than taking exacting notes) 2. Come to lecture • ¼ of class time spent solving problems 3. Come to recitation section 4. Do homework • best way to prepare for exam is by doing problems 5. Don’t fall behind! Cover 1 Chap/week! 6. If you have questions, get them answered 7. Discuss and explore material with classmates 8. Come to office hours when you need help Course Outline Mechanical Waves 14.Waves 15. Superposition and Interference of Waves Electro-Magnetic Waves 34.Maxwell’s Equations 35.Light 36.Mirrors and Lenses and their uses 37.Interference 38.Diffraction Particle Waves 40. Quantization of Angular Momentum and Energy Values 41.Quantum Physics 42.Quantum Effects in Large Systems of Fermions and Bosons Motivation Why learn about waves? 1. Fundamental, quantitative, and creative thinking → what you learn Physics 207 will make you a better scientist or engineer 2. Fundamental to classical and modern physics • How does your TV/radio function? • How do telescopes and microscopes work? • Why don’t hydrogen atoms and neutron stars implode? 3. Material is interesting and exciting (of course…!)! z(x,t) x or t A wave is an organized oscillation causing a displacement in z (position, pressure, temperature, electric field, density, etc) as a function of x or t Waves: • mechanical (sound, water, etc.) • non-mechanical (EM) continuation of Physics 108 • optics • quantum mechanics: electron, photon, hydrogen atom • Schroedinger’s wave equation • Pauli principle • Periodic table • Optical properties of materials • Lasers Mechanical Waves medium: material that carries the wave, e.g., air, water, solid. Can be modeled as masses connected by springs Transverse waves Longitudinal waves Longitudinal and transverse waves a+ b =d c Water is incompressible Traveling wave (speed of wave determined by mass density and tension) The Wave Equation Look at a short segment of a string: length: ∆x mass: m = µ∆x String is under tension T We look at a transverse wave => look at the transverse net force on the segment For small angles: And we also know: T sin θ ≈ T tan θ ∂ tan θ = z ( x, t ) ∂x With this, we can calculate the net force on the String segment ∂ FL = −T sin θ L ≈ −T z ( x, t ) ∂x and ∂ FL = T sin θ R ≈ T z ( x + ∆x, t ) R ∂x The combined force on the string segment is ∂ ∂ F = FR + FL = T [ z ( x + ∆x, t ) − z ( x, t )] ∂x ∂x Recall: the definition of the derivative is d f ( x + ∆x ) − f ( x ) f ( x) = dx ∆x And we get ∂ ∂z ∂2 z F = T ( ∆x ) ( ) = T ( ∆x ) 2 ∂x ∂x ∂x Note that curvature, not displacement determines F! And together with Newton’s 2nd law ∂2 z ma = µ ( ∆x ) 2 ∂t we get ∂2 z ∂2 z F = ma = µ ( ∆x ) 2 = T ( ∆x ) 2 ∂x ∂t And after canceling the ∆x: ∂2 z ∂2 z µ 2 =T 2 ∂t ∂x This is the wave equation for a string The wave equation applies to standing waves and traveling waves! (We assumed neither one during derivation!) In its more general form, the wave equation describes not just harmonic waves on a string, but all forms of harmonic waves Wave equation for a string: ∂2z µ ∂2z = 2 T ∂t 2 ∂x Dimension of µ : T 2 1 µ [M / L] T = = = T ML / T 2 L2 velocity2 [ ] [ Dimension suggests that We define µ T v≡ ] plays the role of a speed T µ We will see that v is the speed with which a wave (or disturbance) propagates ∂2z 1 ∂2z = 2 2 2 v ∂t ∂x General form of the wave equation ∂2z 1 ∂2z = 2 2 2 ∂x v ∂t The wave equation applies not just to strings with mass densities µ and tension T, but to many other waves (like sound, light, radio waves, …) For the string, the speed of waves was v = T µ For mechanical waves, it is in general true that: v= restoring force factor mass factor Demo: speed of wave on latex tube Wave speed depends on medium [ tension ] kg ⋅ m/s 2 m 2 = = 2 = v 2 [lin. mass density] kg/m s restoring force factor v= mass factor v↑ if restoring force ↑ (quicker snap back) if mass factor ↓ (less inertia) Traveling wave latex tube demo: Filled with air Filled with water How else can we change the wave speed? Traveling Waves Consider a disturbance moving along the string: Note that the shape of the wave does not change! What is the most general way to describe this behavior? z ( x, t ) = f ( x − vt ) z v t0=0 0 x z = f ( x) z 0 v 0 → x2 = x1 − vt1 + vt2 t2=2T x2 z = f ( x − 2vT ) x2 = x1 + v ( t2 − t1 ) x → moving to right, x must increase as t3=3T t increases to keep z 0 0 x x2 − vt2 = x1 − vt1 x1 z = f ( x − vT ) z z z ( x,t ) = f ( x − vt ) t1=T x3 x z = f ( x − 3vT ) v t4=4t x4 x z = f ( x − 4vT ) ( x - vt ) constant Simple Harmonic Motion Use sine or cosine as function: f ( x − vt ) ⇒ sin[k ( x − vt )] k, the wave number, makes the argument unitless. z ( x, t ) = z0 sin[k ( x − vt )] = z0 sin[kx − kvt ] = z0 sin[kx − ωt ] where ω = kv z ( x, t ) = z0 sin( kx − ωt ) Travels into the positive x direction z ( x, t ) = z0 sin(kx + ωt ) Travels into the negative x direction Periodic Waves v Wavelength: P λ : Length after which the wave repeats itself 2π (spatial Wavenumber: k= frequency) λ Frequency: number of crests passing point p (temporal f: frequency) 1 second Angular frequency: ω = 2π ⋅ f Period: τ : Time between crests passing point P 1 2π τ= = f ω Speed of a Wave 2π λ= k As we watch a wave moving through, we measure the time ∆T it takes for one wavelength λ to pass. λ is a distance and the wave moved this distance in the time ∆T => The speed of the wave is: v = λ ∆T The time for one wave to pass is called the period τ and we have λ v = = λ ⋅ f because f = 1 τ τ Example: Every 3 seconds a 2 m long wave passes you λ 2m 2 m => v = = = τ 3s 3 s Note that: ω f = 2π and 2π λ= k and therefore 2π ω ω ⋅ = v = λf = k 2π k If we plug in a simple sinusoidal wave into the wave equation: z ( x, t ) = z0 sin( kx − ωt ) 1 ∂ z ∂ z = 2 2 2 v ∂t ∂x 2 2 1 ∂ ∂ (− ω cos(kx − ωt ) ) = z0 (k cos(kx − ωt ) ) z 2 0 v ∂x ∂t 1 2 2 z ω sin( kx − ω t ) = z k sin(kx − ωt ) 0 2 0 v 1 2 2 = k ω v2 z ( x, t ) = z0 sin(kx − ωt ) satisfies the wave equation if v= ω k Traveling wave f ( x − vt ) satisfies the wave eq. y ( x,t ) = x − vton f ( x,t ) = f ( x − vt ) = f ( y ) single variable y replaces x and t want to show that ∂2 f ( y ) 1 ∂2 f ( y ) = 2 2 ∂x ∂t 2 v recall the chain rule for differentiation: ∂f ∂f ∂y ∂f ∂f ∂y = and = ∂x ∂y ∂x ∂t ∂y ∂t ∂f ( y ) df ( y ) ∂y → = = dy ∂x ∂x ∂y = ∂x HW Problem # 15 17 ∂y = ∂t Power delivered by a traveling wave z T θ Tsin θ Tcos θ 1D system x O Power moving past point x: x P = Ftransverse ⋅ vtransverse The transverse force is (for small θ ) Ftransverse ∂z = −T sin θ ≈ −T tan θ = −T ∂x and the transverse velocity of the string is just vtransverse ∂z = ∂t For a harmonic wave z ( x, t ) = z0 sin(kx − ωt ) the power moving past point x is then: ∂z ∂z P = Ftransverse ⋅ vtransverse = −T ∂t ∂x = Tkω z02 cos2 ( kx − ωt ) = µvω z cos ( kx − ωt ) 2 2 0 => 2 P = µvω z cos ( kx − ωt ) 2 2 0 2 and the time average of this is 1 2 2 P = µvω z0 2 (the average over cosine squared over long time is ½) Energy Transport and Energy Density We did not consider any friction or other mechanism for loss of mechanical energy of the wave. ⇒Power supplied by “emitter” is transmitted with the wave ⇒Power is a rate of energy transport Vibrating string has an energy density µ (energy per unit length) ∆l = v ⋅ ∆t P(x,t) v x x0 Energy passing through point x0 in time ∆t is ∆E = P ⋅ ∆t The energy density (energy per unit length) is then Length ∆l = v∆t moves past point x in time ∆t ∆E P∆t P u( x,∆t E ) == P∆t =moves =past point x in time ∆t Energy ∆l v∆t v energy density per2unit u ( x,t ) 2 length 2 = µω z0 cos ( kx − ωt ) ∆E P∆t P u ( x,t ) = = = ∆l energy v∆t perv unit length is The average = µω 2 z02 cos12 ( kx2− ω 2 t) E = µω z0 average energy per2unit length E 1 E = µω 2 z02 → contains both kinetic and 2 potential energy. For a simple harmonic oscillator E = K +U 1 K = E 2 1 U = E 2 28 See problem # 32 Standing Waves wavelength determined by boundary conditions z x Mathematical description of standing waves z ( x,t ) = f ( x ) g ( t ) g ( t ) = cos (ω t + φ ) where is ω the frequency of simple harmonic motion and φ is the phase. Initial conditions determine φ e.g., if g ( t = 0 ) = 1 → φ = 0 if g ( t = 0 ) = 0 → φ = ? → Note that all parts of the wave have the same time-dependence f ( x ) = z0 sin ( kx + δ ) k ≡ wavenumber is the spatial frequency (analogous to temporal frequency ω ) k= 2π Which has a larger k? λ x Rad → number of oscillations*2π per meter [k ] = m z Z0 z(x,t)=g(x)*f(t)=Z0sin(kx)*cos(ωt) t = 0 → cos(0) = 1 x -Z0 Increasing time z Z0 ωt = π/4 → cos(π/4) = 0.71 x -Z0 z Z0 ωt = π/2 → cos(π/2) = 0 x -Z0 z x L Nodes at ends (let δ = 0) sin ( k × 0 ) = 0 and sin ( k × L ) = 0 → k × L = nπ for n = 1, 2, 3, 4, … nπ → kn = L 2π 2 L → λn = = kn n → λ is quantized! ⇒Have found the permitted wave numbers kn = 2π λ = nπ L ⇒This tells us how many half wavelengths fit within the boundaries 2L λn = n There are three half wavelengths => third harmonic ⇒There are TWO nodes ( ) There are always n − 1 nodes in a standing wave of the nth harmonic mode Sill needed: how fast does this wave swing? Found the permitted wave numbers k or the permitted wavelengths n of the standing wave with boundary conditions Now: want to find the angular frequency ω , and thus, how fast the standing wave oscillates. Reminder: Angular frequency ω = 2πf ω is determined by “dynamics”, that is by Newton’s 2nd law: F = ma How to apply this to get ω ?????? Frequency of Standing Waves Substitute our standing wave into the wave equation: z ( x, t ) = z0 sin(kx + δ ) cos(ωt + ϕ ) ∂2z 2 = z sin( kx + δ )( − ω ) cos(ωt + ϕ ) 0 2 ∂t => ∂2 z 2 = z ( − k ) sin( kx + δ ) cos(ωt + ϕ ) 0 2 ∂x ∂2 z ∂2 z and therefore µ 2 = T 2 becomes: ∂t ∂x µω 2 = Tk 2 or ω =k T µ = 2π T λ µ for strings in general, and for standing waves on a string: ωn = nπ L T µ Demo: Standing wave on a string nπ z ( x) = A sin ( kn x ) = A sin x at fixed time L z x L n= fn = n= fn = n= fn = n= fn = Do our results fit to theory? n T fn = 2L µ Summary Standing Waves: Longitudinal and Transverse z ( x,t ) = f ( x ) g ( t ) = z0 sin ( kx + δ ) cos (ω t + φ ) boundary conditions → kn and λn (discrete) properties of medium → ω and τ Traveling Waves: z ( x,t ) = f ( x − vt ) e.g., z0 sin ( kx − ω t ) Both satisfy the wave equation: ∂2 z 1 ∂2 z = 2 2 2 ∂x v ∂t Sound Waves No net flow of molecules in direction of wave! ∆ρ = Aρ sin ( kx − ω t ) Why no transverse waves? Speed of sound in air: γ 0 P0 v= ρ0 γ 0 = 1.4 , v = 330 m/s = 740 mi/hr at STP in solids: v= Y ρ , Y = Young's modulus and ρ = density Y =? F A Y]=? [ ∆L L F stiff material Y ↑ Y= soft material Y ↓ ∆L A L Speed of sound waves in various media Y v= ρ Medium Density Hydrogen Air Liquid mercury (20oC) Methyl alcohol Water Polyethylene Lead Silver Aluminum Beryllium 0.09 g/l 1.3 g/l 11.3 g/cm3 10.5 g/cm3 2.7 g/cm3 1.85 g/cm3 Speed of sound (m/s) 1284 330 1450 1189 1402 920 1210 2700 5000 12870 Pressure Wave Both ends open Displacement Amplitude z(x,t) Pressure Amplitude (also called ∆ρ(x,t)) Standing sound waves Boundary condition: no displacement on closed ends and maximum displacement on open ends displacement v=λf v = 330 m/s (fixed by P0 and ρ0 ) 2L λn = n v n v fn = =v = nf1 where f1 = λn 2L 2L For L = 4.0 m 330 m/s f1 = = 41 Hz 2 ( 4.0 m ) ff32 = = 23 ff11 ==83 124HzHz f 3f 124 H f 2 = 2 f1 = 83 Hz z=0 x f1 ∂z =0 ∂x f2=3f1 f3=5f1 ∂z =0 ∂x f1 x ∂z =0 ∂x f2=2f1 f3=3f1 Demo: The trumpet mouthpiece Mouthpiece + trumpet (cavity) 1. Mouthpiece: frequencies and volume? 2. Mouthpiece+Trumpet: frequencies and volume? 3. Changes in cavity length: what happens to frequency? a) Pressing a valve b) Pulling slide c) Changing bell shape with hand Intensity P I= A Ptotal [I ] = W/m 2 PA Want to keep track of I over many orders of magnitude → use logarithmic scale. This scale deals with the exponent of 10 required to obtain I relative to a very small reference intensity I0. I β ≡ 10 log10 → 10 times the number of orders of I0 [ β ] = dB magnitude that I is larger than I 0 or decibels, I 0 = 10−12 W/m 2 I = I0 → β = 0 0 orders of magnitude I = 10 I 0 → β = 10 1 order of magnitude I = 100 I 0 → β = 20 2 orders of magnitude I = 105 I 0 → β = 50 5 orders of magnitude human ear can hear 0-120 dB range (10-12 − 1 W/m 2 ) range for human eye is similar (though I0 is different) The Doppler Effect Moving source, stationary observer v vs λ= f0 v v A Source moving away from observer A lengthened λ Source moving towards v v + vs λ' = λ + s = f0 f0 vs v − vs λ' = λ − = f0 f0 observer not moving, sees waves travel at v observer not moving, sees waves travel at v v v f'= = f0 v v λ' + s f ' < f0 v v = f0 f'= − λ' v v s f ' > f0 B observer B shortened λ source moving away from source moving towards observer (v opposite sign of vs ) observer (v same sign of vs ) Demo: Doppler effect due to moving source and stationary observer vs Observer Doppler effect: stationary source, moving observer Since source is not moving, wavelength in medium is unchanged → λ’=λ v vr v v v However, the effective velocity of the sound wave changes because the observer is moving with respect to the medium! v vr v v v λ' = λ λ' = λ observer moving away from observer moving towards source, sees waves travel at v' source, sees waves travel at v' v' = v + vr v' = v − vr f'= v' λ = v − vr λ f'= v' λ = v + vr λ v f ' = f 1 − r v f ' < f0 v f ' = f 1 + r v f ' > f0 observer moving away from source (v same direction as vr ) observer moving towardssource (v opposite direction as vr ) Doppler effect: both source and observer moving 1. Start with source, which is moving with respect to the medium at a speed vs λ' = λ' v − vs → wavelength in medium f0 vs v λ' v 2. Observer moving at speed vr with respect to medium, sees λ ' and new wave speed v' v' = v − vr (no longer need source) λ' v λ' v − vr v' v − vr = = f'= f0 λ ' ( v − vs ) f 0 v − vs vr v Note that f’ is not symmetric! Receiver moving at 5 km/hr towards a stationary source is NOT the same as having the source moving at 5 km/hr towards a stationary observer! Sign conventions for Doppler calculation Level 1: a) If vs (vr) is in the same direction as v, use positive value for vs (vr) . b) If vs (vr) is in the opposite direction as v, use negative value for vs (vr) . Plug into formula v − vr f' = f0 v − vs Level 2: Choose signs so that : a) source moving towards observer increases f’ b) observer moving towards source increases f’ Level 3: a) Start at source and calculate λ’ b) Look at receiver to get relative wave speed v' v’ f'= λ' Traveling waves generated at a fixed point (X), in a moving medium X vm Strea m flo vs=? vr=? w v=speed of sound 90o-θ1 not θ1 vt2 v = cos θ 0 = vs t2 vs v ( t2 − t1 ) v = cos θ1 = vs ( t2 − t1 ) vs vs Mach number ≡ v Large Mach number : θ1 → 90" vs < v vs > v Shock wave produced by supersonic bullet passing over a candle Charged particle moving through water at a speed greater than the speed of light in water → Cerenkov radiation Ch 14 Useful Equations Standing wave: z ( x,t ) = sin ( kx + δ ) cos (ω t + φ ) Transverse wave on string: ∂2 z ∂2 z T 2 =µ 2 ∂x ∂t 2π T ω= λ µ Wave equation: ∂2 z 1 ∂2 z = ∂x 2 v 2 ∂t 2 Sinusoidal traveling wave in x-direction: z ( x,t ) = z0 sin ( kx − ω t ) v=λf = λ ω T = = τ k µ Energy density: dE = µω 2 z02 cos 2 ( kx − ω t ) dx Average power delivered by wave: 1 P = µω 2 z02 v 2 γ P0 vsound = ρ0 I I0 β = 10 log10 λ' = v − vs f0 f'= v ' v − vr = f0 λ ' v − vs where I 0 = 10−12 W/m 2 v' = v − v r