Traveling Harmonic Waves

Transcription

1 Traveling Harmonic Waves 6 January 2016 PHYC 1290 Department of Physics and Atmospheric Science

2 Functional Form for Traveling Waves We can show that traveling waves whose shape does not change with time have a special functional form. Consider the following snapshot graph: f(x) t = 0 s Let y = f(x) describe the initial displacement.

3 Suppose at some later time the wave has traveled distance Δx : Δx f(x-δx) t = 15 s The new displacement relation is y = f(x-δx) We can write Δx = vt, where v is the phase speed and t is the time. Substituting into the displacement relation yields: y = f(x-vt) Note: Waves with v > 0 travel to the right; v < 0 travel to the left.

4 Traveling Harmonic Waves Using the functional form for traveling waves, we can write a general form for traveling harmonic (i.e., sinusoidal) waves as y = A sin[k(x vt)] Multiplying through gives y = A sin(kx kvt) or y = A sin(kx t) where = kv The amplitude ; same units as y The wavenumber ; units [length -1 ] Angular frequency ; units [time -1 ] Phase speed ; units [length/time]

5 Q: How are k and ω related to wavelength and period? We can determine this by using history and snapshot graphs.

6 Let s plot a history graph for y = sin(kx - ωt) at x = 0. y(x = 0) = sin(-ωt) = -sin(ωt) t y = -sin(ωt) y One period T 0 0 π/2ω -1 0 π/2ω π/ω 3π/2ω 2π/ω 5π/2ω t π/ω 0 3π/2ω 1 2π/ω 0 5π/2ω -1 T = 2π/ω or ω = 2π/T Substituting the definition of standard frequency f = 1/T yields ω = 2πf.

7 Now, let s plot a snapshot graph for y = sin(kx - ωt) at t = 0. y(t = 0) = sin(kx) x y = sin(kx) y One wavelength λ 0 0 π/2k 1 π/k 0 0 π/2k π/k 3π/2k 2π/k 5π/2k x 3π/2k -1 2π/k 0 5π/2k 1 λ = 2π/k or k = 2π/λ

8 Q: How are wavelength and frequency related? We wrote before that ω = kv. v = ω/k Substituting ω = 2πf and k = 2π/λ gives v = fλ The equation for v (either version) is sometimes called the fundamental relationship for sinusoidal waves.

9 Particle Speed and Acceleration Consider a harmonic wave described by y = Asin(kx - ωt) What are the speed and acceleration of the particle at x = 0? Like before, y(x = 0) = -Asin(ωt) The speed of the particle at x = 0 is v y = dy dt = A cos( t) and acceleration is a y = dv y dt = A 2 sin( t)

10 Notice that the speed and acceleration depend on time. What are the speed and acceleration of the same particle at t = 0? v y (t = 0) = A cos[ (0)] = A the particle is moving downward at t = 0 a y (t = 0) = A 2 sin[ (0)] = 0 the particle is not accelerating at t = 0 and must therefore be at its maximum absolute speed

11 Do the particle speed and acceleration results make sense? Consider the snapshot graph at t = 0 s: y t = 0 x As the wave moves to the right, the particle will move downward (as predicted). The particle must have 0 speed at the crest and the trough, and so will be a maximum in between (as predicted). So yes, the results make sense.

12 Extra Material

13 Function Translation Consider the following snapshot graphs: y f(x) t = 0 s x1 x y Δx g(x) t = 15 s x1 x2 x Q: How are g(x) and f(x) related?

14 A: Consider for the wave crests that g(x2) = f(x1) Since the wave travels with phase speed v, we can write x2 = x1 + Δx Re-arranging and substituting gives g(x2) = f(x2-δx) Generalizing for any point on the wave yields g(x) = f(x-δx) Thus, a translation of f(x) to the right by an amount Δx is given by f(x-δx), as used in the lecture.

15 Wave Function Example Consider the following function y = 0.5 sin(2x t) x 0.5t Does the function describe a traveling wave? YES. The function can be rewritten as y =0.5 sin[2(x x 0.5t 0.5t)] which is of the form y = f(x - vt) with v = 0.5 m/s. Note: There can be only one phase speed v for a traveling wave!

16 What is the speed of the particle at x = 0 m at t = π/2 s? Assume that y is measured in cm. v y (x = 0, t) = d dt y(x = 0, t) = d dt Simplifying while noting that sin(-x) = -sin(x) gives v y (x = 0, t) = d t 1 sin(t) dt ( = 0 ) = sin( ) + cos Applying the product rule of differentiation, d da (AB) = dt dt B + AdB dt dt v y (x = 0, t) = sin[2(0) t] 0.5 (0) 0.5t t 2 sin(t) + t 1 cos(t) 4 v y (x = 0, t = /2) = 4 2 cm/s

17 Harmonic Wave Example Suppose a 25 m long water wave travels at 5 m/s and has a trough-to-crest height of 2 m. a) What is the equation for the wave? k = 2π/λ = 2π/(25 m) = 0.25 m -1 and A = 1 m y = sin[0.25(x-5t)] = sin(0.25x-1.25t) b) How often will a stationary buoy be lifted by wave crests? T = 2π/ω = 2π/(1.25 s -1 ) = 5 s every five seconds.

18 Why is k called the wavenumber? We know that wavenumber k is related to wavelength λ by k = 2π/λ The wavenumber k is therefore the number of waves in 2π length. For example, consider a wave with λ = π/2 m: y 4 cycles 0 2π x (m) k = 2π/λ = 2π/(π/2 m) = 4 m -1

19 Wave Speed on a String: Derivation

20 Consider this snapshot of a sinusoidal wave on a string at t = 0: y = A cos(kx) y 0 x The acceleration of the particle on the string at x = 0 is ay = -Aω 2 cos(ωt) At t = 0 this gives a y = A 2

21 The acceleration is governed by Newton s Second Law F y = ma y so the force on a particle with mass m at the crest of the wave is given by F y = ma! 2 We can derive the force another way and compare.

22 Considering the tension on the segment of length Δx around a particle on a string as shown: y θ T sin T T x -Δx/2 0 Δx/2 Tension vectors may be drawn as tangents to the string at each end of the segment. The transverse force on a segment is therefore F y = 2T sin

23 Comparing the two equations for force gives ma 2 = 2T sin Since the angle θ is very small, we can write sin tan Δx/2 θ dx dy T Now tan θ is just the slope of the tangent at x = Δx/2; i.e., tan = dy Therefore ma 2 = dx x= x/2 2T dy dx x= x/2

24 Substituting the waveform y = A cos(kx) gives ma 2 = 2T d dx A cos(kx) x= x/2 = 2T Ak sin(kx) x= x/2 = 2T Ak sin(k x/2) For small segments we can make the approximation sin(k x/2) = k x/2 and so ma 2 = T Ak 2 x

25 Rearranging gives k 2 = T m/ x Suppose the string has a linear density defined by μ = m/l where m is the string s mass and L is its length. This allows us to write m = µδx for the segment. Substituting this and v = ω/k yields v = T µ