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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)

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Physics diagram aids