Travelling and Standing Waves
Many 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 major
component of this biophysics course.
Sound Waves:
(Acoustics, vibrating strings)
Study guide 2
Electromagnetic Waves:
(Light, X-rays, Infra-red)
Study guide 3-4
Quantum Mechanical Waves:
Study guide 4-6
(motion of elementary particles/waves)
1
MATERIAL TO READ
Web:
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 waves
Handbook: 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 waves
3. How to graph a wave given the equation
4. Relations between period (T), frequency (f or
υ) and wavelength (λ)
Type of wave Form
Sinusoidal
y = Asin(x)
Square
y = A for 0 < x < λ/2,
= -A for λ/2 ≤ x ≤ λ
Ramp
y = Cx, for -λ/2 ≤ x ≤ λ/2
Periodic functions can be described as a sum of
sinusoidal waves. Therefore we need trigonometry
2
Trigonometry Review
Trig. Functions defined on a circle…
r
θ
O x
P
s
y
θ
Point P moves around
a circle of radius r
s
(arc length)
θ = s/r
dimensionless unit:
“ radians “
See Web tutorial
to review dimensions
Common unit is “degrees”:
Full circle = 360 o
For full circle, s = circumference = 2πr
∴ s/r = 2π
π = 360o
⇒ 1 radian = 360o/(2π
π) = 57.3o.
3
Definitions:
sin(θ)
Cos(θ)
tan(θ)
y/r
x/r
y/x
Opposite/hypotenuse
Adjacent/hypotenuse
Opposite/adjacent
SOH
CAH
TOA
Be careful calculating these functions using
calculator: “DEG” or “RAD” mode on calculator
indicates units.
No degree symbol
“o” means angle is in
Examples: sin(40.0o) = 0.642
radians
cos(2.5) = -0.801
o
π = 360
π = 180
o
−π/2 = -90
1
sin(θ)
π/2 = 90
o
2
o
As P moves around circles sin(θ) and cos(θ) vary…
0
-1
-2
-6
-4
-2
0
2
θ
4
6
8
10
4
o
π = 360
o
π = 180
o
−π/2 = -90
1
cos(θ)
π/2 = 90
o
2
0
-1
-2
-6
-4
-2
0
2
4
6
8
10
θ
NOTES:
1. Positive angles are measured counterclockwise
from positive x axis
2. Both sinθ and cosθ repeat once every time P
rotates around circle (2π radians)
i.e. for any integer n,
sin(θ+2πn) = sin(θ)
Periodicity
cos(θ+2πn) = cos(θ)
3. Also:
sin(-θ) = -sin(θ)
ODD
cos(-θ) = cos(θ)
EVEN
5
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’s
say it is called sin-1. Let’s apply this function to
both 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
(R: radians)
(D: degrees)
sin-1(0.683) = 0.752
= 43.1o
Due to the periodic nature of sin, cos and tan
there are actually an infinite set of angles that
would produce the same value of each function…
2
sin(θ)
1
f
0
NB
-1 < f < 1
-1
-2
0
2
4
6
8
10
θ
All 4 values of θ have same f; only 2 per revolution.
7
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 can
be 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 θ IS NEGATIVE
Flip the above diagram around to see it.
Example θ = sin-1(-0.48)…
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Inverse Cosine:
θ1
2
θ2
cos(θ)
1
0
-1
-2
0
2
4
6
θ
Example:
θ = cos-1(0.45) = 63.3o
What is other value?
y
x1
θ2
θ1
x2
x
x1 = cos(θ1) = cos(θ2) = x2
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From the diagram we see that the other value for
θ is, θ2 = 360o - θ1 = 296.7o
= -θ1
NB!
If we flip the above diagram, we see that this also
works 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:
2
tan(θ)
1
θ1
θ2
0
-1
-2
-2
0
2
4
6
θ
Example:
θ = tan-1(1.0) = 45.0o
What is other value?
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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 for
θ is, θ2 = θ1 – 180o = -135.0o = 225.0o.
If we flip the above diagram, we see that this also
works 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 practice
on…
11
Oscillations and Travelling Waves
e.g. hanging weight on spring
y
y=0
Equilibrium
Position
y = y0
Initial
position
Spring stretched to initial position, y0, and
released. The mass oscillates between y0 and -y0 in
"Simple Harmonic Motion" or SHM.
Equation of motion:
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ΣF = ma ⇒
⇒
-ky = may
-(k/m)y = (d2y/dt2)
Let's try:
y(t) = y0sin(ωt)
where
and
(Ideal Spring Approx.)
⇒
Diff. Equation
dy/dt = ωy0cos(ωt)
ω = 2π
π/T = angular frequency
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 ν = number of cycles or
oscillations per unit time (usually seconds).
e.g.: Let, T = 0.5 s (period)
in one second 2 oscillations occur
that is: ν = 2 cycles/second
υ = 1/T = ω/2π
π
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 is
time 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 sinusoidal
λ
y0
x
-y0
λ
wave repeats itself after distance, λ = wavelength.
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y as a function of t at any fixed position is also
sinusoidal
T
y0
x
-y0
T
In one period T, wave moves forward 1 wavelength.
speed of Wave = Distance/time = λ/T = λν
∴v = λν
governs all wave motion
(water ripples, sound, light…)
for velocity must also provide direction along +ve
or –ve x.
Equations for Travelling Wave (Function of t, x)
y = y0sin(ωt ± kx) “+” or “-” according to direction
where k = 2π
π/λ
λ = wave number
y = y0sin(2πt/T ± 2πx/λ)
y = y0sin(2πνt ± kx)
y = y0sin(k(vt ± x))
since kv = (2π/λ)v = 2πν = ω
16
If we look a t a particular position, equation
reduces: e.g. x = 0 ⇒ y = y0sin(2πt/T).
y
3T/4
T/4 T/2
2T
t
T
Alternatively can take snapshots at particular
time, equation reduces again: y = y0sin(-2πx/λ)
(we have assumed +ve x velocity for wave)
y
λ/4 λ/2
λ
3λ
λ/4
2λ
λ
x
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 of
crest) vs. t.
π/2 - 2πt/T = ± 2πx/λ
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± x = λ/4 - λt/T
± x = λ/4 - vt
since λ/T = λν = v
This separates into two equations:
Upper sign
x = λ/4 – vt
-ve velocity
Lower sign
and - x = λ/4 – vt
⇒
x = vt - λ/4
+ve velocity
Therefore, in the equation:
y = y0sin(2πt/T ± 2πx/λ)
The upper sign corresponds to wave crest
velocities in the negative x direction and the lower
sign to those in the positive x direction.
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Sample Problem
example 1-5 from the text
A wave moves along a string in the +x direction
with a speed of 8.0 m/s, a frequency of 4.0 Hz and
amplitude 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 sign
given by fact that wave travels in +ve x
direction (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
λ
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
20
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
4
y (m)
1/6
1/3
-4
Redraw using x = x’ + 0.45
y (m)
1/6+0.45
1/3+0.45
= 0.62 m
= 0.78 m=
x’ (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 (modified
from 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
4
y (m)
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 move
e.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 waves
with 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 the
sum 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 a
STANDING WAVE
Product of a time dependent term, (cos(ωt)), and a
position dependent term, sin(kx).
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y can be expressed as:
2y0
y
y = a(t)sin(kx),
where a(t) = 2y0cos(ωt)
loop = segment between nodes
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
25
Sample Problem
An incident wave
y1 = 3sin(2t - 3πx) and
a 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 shown
previously 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:
b)
y = -6cos2t sin3πx
What are the…
standing wave’s
(i)
(ii)
(iii)
(i) amplitude
(ii) period…
(iii) and wavelength?
amplitude = 2y0 = 6 m.
ω = 2 s-1 = 2π/T ⇒ T = π s
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 = 0
y = -6cos2t sin3πx ⇒ y = -6cos2(0) sin3πx
= -6sin3πx
6
y (m)
1/3
2/3
x (m)
-6
=0
At t = T/4 = π/4 s
y = -6cos2t sin3πx ⇒ y = -6cos(2π/4) sin3πx
y = 0 for all x
6
-6
y (m)
1/3
2/3
x (m)
=-1
At t = T/2 = π/2 s
y = -6cos2t sin3πx ⇒ y = -6cos(2π/2) sin3πx
y = 6 sin3πx (opp. of t = 0 graph)
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Physics diagram aids
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