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Travelling and Standing WavesMany biological phenomena are cyclic.
• Heartbeats• Circadian rhythms,• Estrus cycles• Many more e.g.’s
Such events are best described as waves.
Therefore the study of waves is a majorcomponent of this biophysics course.
Sound Waves: Study guide 2(Acoustics, vibrating strings)
Electromagnetic Waves: Study guide 3-4(Light, X-rays, Infra-red)
Quantum Mechanical Waves: Study guide 4-6(motion of elementary particles/waves)
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MATERIAL TO READWeb:http://www.physics.uoguelph.ca/biophysics/www.physics.uoguelph.ca/tutorials/GOF/
… /tutorials/trig/trigonom.html… /tutorials/shm/Q.shm.html
Text: chapter 1, Vibrations and wavesHandbook: study guide 1
For 1st quiz, you should know:
1. How to evaluate sin(x), cos(x), sin-1(x), cos-1(x)2. Equations for travelling and standing waves3. How to graph a wave given the equation4. Relations between period (T), frequency (f or
υ) and wavelength (λ)
Type of wave FormSinusoidal y = Asin(x)Square y = A for 0 < x < λ/2,
= -A for λ/2 ≤ x ≤ λRamp y = Cx, for -λ/2 ≤ x ≤ λ/2
Periodic functions can be described as a sum ofsinusoidal waves. Therefore we need trigonometry
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Trigonometry Review
Trig. Functions defined on a circle…
P Point P moves aroundr s a circle of radius r θ y
O x
s θ (arc length)
θθ = s/r
dimensionless unit: See Web tutorial “ radians “ to review dimensions
Common unit is “degrees”:Full circle = 360 o
For full circle, s = circumference = 2πr∴∴ s/r = 2ππ = 360o
⇒⇒ 1 radian = 360o/(2ππ) = 57.3o.
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Definitions:
sin(θ) y/r Opposite/hypotenuse SOHCos(θ) x/r Adjacent/hypotenuse CAHtan(θ) y/x Opposite/adjacent TOA
Be careful calculating these functions usingcalculator: “DEG” or “RAD” mode on calculatorindicates units.
Examples: sin(40.0o) = 0.642
cos(2.5) = -0.801
As P moves around circles sin(θ) and cos(θ) vary…
No degree symbol“o” means angle is inradians
θ-6 -4 -2 0 2 4 6 8 10
sin(
θ)
-2
-1
0
1
2
π/2
= 9
0o
π =
180
o
π =
360
o
−π/2
= -
90o
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NOTES:
1. Positive angles are measured counterclockwisefrom positive x axis
2. Both sinθ and cosθ repeat once every time Protates around circle (2π radians)
i.e. for any integer n,sin(θ+2πn) = sin(θ)cos(θ+2πn) = cos(θ)
3. Also:sin(-θ) = -sin(θ) ODDcos(-θ) = cos(θ) EVEN
θ-6 -4 -2 0 2 4 6 8 10
cos(
θ)
-2
-1
0
1
2
π/2
= 9
0o
π =
180
o
π =
360
o
−π/2
= -
90o
Periodicity
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Small Angle Approximation:
For θ < π/20 radians ≅ 10o:θ MUST BE IN RADIANS
• sinθ ≅ θ• cosθ ≅ 1• tanθ = sinθ/cosθ ≅ θ
Not used until Study Guide 4.
Inverse Trig. Functions: Inverse Sine
sinθ = f (given f)
what is θ?
Imagine there is an inverse function for sin. Let’ssay it is called sin-1. Let’s apply this function toboth sides of the equation .
sin-1(sin(θ)) = sin-1(f) ⇒ θ = sin-1(f)
LOOK!THERE IS SIN-1 ON YOUR CALCULATOR
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• sin-1 also called “arcsin(f)”• similar functions for cos-1(f) and tan-1(f).
Example:Calc. setting
sin-1(0.683) = 0.752 (R: radians)= 43.1o (D: degrees)
Due to the periodic nature of sin, cos and tanthere are actually an infinite set of angles thatwould produce the same value of each function…
f
NB-1 < f < 1
All 4 values of θ have same f; only 2 per revolution.
θ0 2 4 6 8 10
sin(
θ)
-2
-1
0
1
2
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Therefore sin-1(f) is defined on interval 0 ≤ f≤ 2π.The two values are : θ AND (π - θ) = (180o - θ).
The reason there are two values per revolution canbe understood with the aid of a diagram.
y
θ2
y1 y2
θ1
x
y1 = sin(θ1) = sin(θ2) = y2
Example: If sinθ = 0.45, θ = ?…
Calculator returns θ = sin-1(0.45) = 26.7o
Another value is: 180o – 26.7o = 153o
THIS ALSO WORKS IF θθ IS NEGATIVE
Flip the above diagram around to see it.
Example θ = sin-1(-0.48)…
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Inverse Cosine:
Example: θ = cos-1(0.45) = 63.3o
What is other value?y x1
θ2
θ1
x x2
x1 = cos(θ1) = cos(θ2) = x2
θ0 2 4 6
cos(
θ)
-2
-1
0
1
2θ1 θ2
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From the diagram we see that the other value forθ is, θ2 = 360o - θ1 = 296.7o = -θ1
NB!If we flip the above diagram, we see that this alsoworks for θ < 0.
Therefore cos-1(f) is also defined on interval 0 ≤ f≤2π. Its 2 values are : θ AND (2π - θ) = (360o - θ).
Inverse tangent:
Example: θ = tan-1(1.0) = 45.0o
What is other value?
θ-2 0 2 4 6
tan(
θ)
-2
-1
0
1
2
θ1 θ2
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y
θ2
θ1
x
x1/y1 = tan(θ1) = tan(θ2) = x2/y2
From the diagram we see that the other value forθ is, θ2 = θ1 – 180o = -135.0o = 225.0o.
If we flip the above diagram, we see that this alsoworks for θ < 0.
Therefore tan-1(f) is also defined on interval 0 ≤ f≤ 2π. Its 2 values are : θ AND (θ - π) = (θ-180o).
See Study Guide for more examples to practiceon…
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Oscillations and Travelling Waves
e.g. hanging weight on spring
y
y = 0
EquilibriumPosition y = y0
Initial position
Spring stretched to initial position, y0, andreleased. The mass oscillates between y0 and -y0 in"Simple Harmonic Motion" or SHM.
Equation of motion:
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ΣF = ma ⇒ -ky = may (Ideal Spring Approx.)
⇒ -(k/m)y = (d2y/dt2) Diff. Equation
Let's try:y(t) = y0sin(ωt) ⇒ dy/dt = ωy0cos(ωt)
where ωω = 2ππ/T = angular frequency and T = period = time for one cycle.
NB: y(t) = y0sin(2π(t/T)) repeats every T
Argument MUST be in radians
⇒ d2y/dt2 = -ω2y0sin(ωt)⇒ d2y/dt2 = -ω2y(t)
∴y(t) = y0sin(ωt) is solution to diff. eq. if ω2 = k/m.
• This is why SHM is sinusoidal.• Other e.g.'s: sound, light, water surface ripples.• Looks like:
T y0
time t -y0
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Other useful definitions:
Frequency = f , or νν = number of cycles oroscillations per unit time (usually seconds).
e.g.: Let, T = 0.5 s (period)in one second 2 oscillations occurthat is: ν = 2 cycles/second
υυ = 1/T = ωω/2ππ
Dimensions of ν: (time)-1
Unit: s-1
or more explicitly, "cycles per second" ≡ Hertz
So we can write equivalent representations:y(t) = y0sin(ωt)
y(t) = y0sin(2πt/T)
y(t) = y0sin(2πνt)
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Travelling Waves
e.g.: ripples on surface of water. Displacement istime and position dependent. Let’s take“snapshots” of y vs. x at three successive times:
(earliest, middle, latest) peaks moving towards right
x(position)
each point moves ONLY vertically
y as a function of x at any fixed time is sinusoidalλλ
y0
x -y0
λλwave repeats itself after distance, λλ = wavelength.
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y as a function of t at any fixed position is alsosinusoidal
T y0
x -y0
TIn one period T, wave moves forward 1 wavelength.
speed of Wave = Distance/time = λ/T = λν
∴∴v = λλνν governs all wave motion(water ripples, sound, light…)
for velocity must also provide direction along +veor –ve x.
Equations for Travelling Wave (Function of t, x)
y = y0sin(ωt ± kx) “+” or “-” according to directionwhere k = 2ππ/λλ = wave number
y = y0sin(2πt/T ± 2πx/λ)
y = y0sin(2πνt ± kx)
y = y0sin(k(vt ± x)) since kv = (2π/λ)v = 2πν = ω
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If we look a t a particular position, equationreduces: e.g. x = 0 ⇒ y = y0sin(2πt/T).
y 3T/4 2T
t T/4 T/2 T
Alternatively can take snapshots at particulartime, equation reduces again: y = y0sin(-2πx/λ)(we have assumed +ve x velocity for wave)
y λλ/4 λλ/2 λλ
x 3λλ/4 2λλ
Wave direction? Crest of wave ⇒ y = y0
y = y0sin(2πt/T ± 2πx/λ)⇒ 1 = sin(2πt/T ± 2πx/λ)⇒ nπ + π/2 = (2πt/T ± 2πx/λ) (for n = 0, 1,2,…)
Let’s use n = 0 case and solve for x (position ofcrest) vs. t.π/2 - 2πt/T = ± 2πx/λ
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± x = λ/4 - λt/Tsince λ/T = λν = v
± x = λ/4 - vt
This separates into two equations:
Upper sign Lower signx = λ/4 – vt and - x = λ/4 – vt
⇒ x = vt - λ/4
-ve velocity +ve velocity
Therefore, in the equation:
y = y0sin(2πt/T ± 2πx/λ)
The upper sign corresponds to wave crestvelocities in the negative x direction and the lowersign to those in the positive x direction.
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Sample Problem example 1-5 from the textA wave moves along a string in the +x directionwith a speed of 8.0 m/s, a frequency of 4.0 Hz andamplitude of 0.050 m. What are the…
(a) wavelength? (b) wave number?(c) period? (d) Angular frequency?(e) Determine an equation for this wave (and
sketch it for t = 0 and t= 1/16 s)(f) What is value of y for the point x = ¾ m a t
time t = 1/8 s?
Solution(a) v = νλ ⇒ v/ν = λ = 8.0/4.0 = 2.0 m
(b) k = 2π/λ = 2π/2 = π m-1
(c) T = 1/ν = 1/4.0 = 0.25 s
(d) ω = 2πν = 8π s-1
(e) y = y0sin(ωt ± kx) = 0.05 sin(8πt - πx) -ve signgiven by fact that wave travels in +ve xdirection (see previous page. For t = 0,y = 0.05 sin(- πx) = - 0.05 sin(πx) :
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y (m) 1.0 2.0
x (m)-0.05
λλ
for t =1/16 s,y = 0.05 sin(π/2 - πx) = - 0.05 sin(π(x – ½))⇒ y = -0.05 sin(πx’), same as above except
x’ = x - ½. y (m)
1.0 2.0 x’ (m)
-0.05
redraw horizontal axis using x’ + ½ = x.
y (m) 0.5 1.5 2.5
x (m)
(f) y =0.05 sin(8πt - πx), for t =1/8 s and x = ¾ m.y =0.05 sin(π - 3π/4) = 0.05 sin(π/4) = 0.035 m
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Sample Problem Self-test III, #1 c
Sketch a graph for y = 4 sin(3πt - 6πx) at t =0.900 s.
y = 4 sin(3π(0.900) - 6πx) = 4 sin(2.7π - 6πx) == -4 sin(6πx - 2.7π) = -4 sin(6π(x – (2.7/6)π))
y = -4 sin(6π(x – 0.45π)) = -4 sin(6πx’)x’ = x – 0.45
k = 6π = 2π/λ ⇒ λ = 2π/6π = 1/3 m
y (m)4 1/6 1/3
x’ (m) -4
Redraw using x = x’ + 0.45y (m) 1/6+0.45 1/3+0.45
= 0.62 m = 0.78 m= x (m)
x = 0.45 m
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Sample Problem Self-test III, #1 d
Sketch y = 4 sin(3πt - 6πx) at x = 0.5 m (modifiedfrom x = 0.4 m).
Since x is fixed ⇒ plot y vs. t
⇒ y = 4 sin(3πt - 6π(0.5)) = 4 sin(3πt - 3π)
⇒ y = 4 sin(3π(t-1)) = 4 sin(3πt’)t’ = t – 1.0
ω = 3π = 2π/T ⇒ T = 2π/3π = 2/3 s
y (m)4 1/3 2/3
t’ (s) -4
Redraw using t = t’ + 1.0
4/3 5/3 t (s)
x = 1.0 m
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Standing Waves
Wave crests do not movee.g.: Vibrating string, sound in organ pipes
Consider first:Reflection of travelling wave from surface
x
Incident wave
Wave inverts upon reflection
Reflected wave
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A standing wave is the sum of two travelling waveswith opposing velocities.
Incident wave (towards –x) y1 = y0sin(ωt + kx)
Reflected wave (towards +x) y2 = - y0sin(ωt - kx)
Inversion of wave
Superposition principle: Resultant wave is thesum of the incident and reflected wave.
ytot = y1 + y2 = y0[sin(ωt + kx) – sin(ωt - kx)]
Use trig. identity: sin(α ± β) = sinαcosβ ± cosαsinβ
ytot = y = y0[ sin(ωt)cos(kx) + cos(ωt)sin(kx)- sin(ωt)cos(kx) + cos(ωt)sin(kx)]
0
y = 2y0[cos(ωt)sin(kx)] Equation for aSTANDING WAVE
Product of a time dependent term, (cos(ωt)), and aposition dependent term, sin(kx).
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y can be expressed as: y = a(t)sin(kx),where a(t) = 2y0cos(ωt)
y loop = segment between nodes2y0
x
-2y0
node antinode
Sketch of standing wave, y vs. x at different times
a(t) > 0
a(t) < 0
At all times, node positions (y = 0) are at same x
At all times, antinode positions are at same x
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Sample ProblemAn incident wave y1 = 3sin(2t - 3πx) anda reflected wave y2 = -3sin(2t + 3πx)produce a standing wave (x, y, t in SI units).
a) Write the equation of the standing wave.Using the summation identity as shownpreviously we can write the following:
General Prescription:Incident wave: y1 = y0sin(ωt - kx)Reflected wave: y2 = - y0sin(ωt + kx)
⇓⇓Standing wave: y = -2y0cosωtsinkx
In this case we obtain: y = -6cos2t sin3πx
b) What are the… (i) amplitudestanding wave’s (ii) period…
(iii) and wavelength?(i) amplitude = 2y0 = 6 m.(ii) ω = 2 s-1 = 2π/T ⇒ T = π s(iii) k = 3π m-1 = 2π/λ ⇒ λ = 2/3 m
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c) Sketch the standing wave at times t = 0 s,t = T/4 and t = T/2.
At t = 0y = -6cos2t sin3πx ⇒ y = -6cos2(0) sin3πx
= -6sin3πx
y (m)6 1/3 2/3
x (m) -6
= 0At t = T/4 = ππ/4 sy = -6cos2t sin3πx ⇒ y = -6cos(2π/4) sin3πx
y = 0 for all x
y (m)6 1/3 2/3
x (m) -6
=-1At t = T/2 = ππ/2 sy = -6cos2t sin3πx ⇒ y = -6cos(2π/2) sin3πx
y = 6 sin3πx (opp. of t = 0 graph)