1.15 Linear superposition applied to two harmonic waves

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1.15 Linear superposition applied to two harmonic
waves

• Can use the idea of linear superposition to
  understand
• Standing waves
• Beats in sound
• Interference effects in light
• Will consider two harmonic waves, y1(x,t) and
  y2(x,t) of equal amplitude A.
• y1(x,t) Asin(k1x-w1t ?1)
• y2(x,t) Asin(k2x-w2t ?2)

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1.15 Linear superposition applied to two harmonic
waves

• From principle of linear superposition we have
• yt(x,t) y1(x,t) y2(x,t)
• yt(x,t) Asin(k1x-w1t ?1) Asin(k2x-w2t ?2)
• To simplify let
• F1 k1x - w1t ?1 and F2 k2x - w2t ?2
• So yt(x,t) Asin(F1) Asin(F2)
• This can be written as

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1.15 Linear superposition applied to two harmonic
waves

• For two harmonic waves interacting at the same
  point and time the resulting displacement is

• From this equation it is evident that the phase
  relationship between the two waves affects the
  resulting displacement.
• Understanding the phase relationship explains
• Standing waves
• Beats in sound
• Interference in light

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1.16 Standing waves on a string

• Consider a perfect string. Two harmonic waves are
  excited on the string, each with the same
  amplitude A, propagating with same frequency w.
  One wave travels left to right, the other right
  to left.
• Here F1 kx - wt ?1 and F2 kx wt ?2
• So
• F1 F2 2kx (?1 ?2)
• F1 - F2 -2wt (?1 - ?2)

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1.16 Standing waves on a string

• If we let ?1 ?2 0 we get

• From this equation it can be seen that the time
  and space coordinates have been separated.
• The time and space coordinates now influence the
  displacement of the wave independently.
• This is the equation for a standing wave

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---Resultant
Initially the two waves are in phase. So from
principle of linear superposition the two waves
add together to give maximum displacement.
Time advances and the two waves move in opposite
directions. A phase difference occurs between the
waves so the displacement due to each individual
wave is different.
Waves continue to move away from each other. The
phase difference increases so net displacement is
reduced.
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Waves continue to move away from each other. The
phase difference increases reaches p so peak of
one wave overlaps trough of the other wave. The
net displacement is zero.
---Resultant
The two waves keep moving opposite and the phase
difference increases so the peak and trough no
longer overlap. Hence the resulting displacement
increases.
Waves continue to move away from each other and
reach a point where the phase difference is 2p
again leading to a maximum in the displacement
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Key Features of Standing Waves
Although the component waves move in opposite
directions the resultant standing wave pattern
does not move. The peak highlighted with the line
is displaced up and down owing to the particle
motion but the peak does not move sideways.
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Key Features of Standing Waves

• There are points where the displacement of the
  standing wave is always zero irrespective of the
  time evolution.
• These points are known as NODES (no
  displacement).
• The nodes occur when
• sin(kx) 0
• This is satisfied when
• kx np (n an integer 0,1,2.)

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Why are nodes formed?

• Let us consider the displacement due to the
  individual traveling waves waves at a node in the
  standing wave pattern.
• At a node, irrespective of the phase difference
  between the two component waves, the displacement
  caused by one wave is equal and opposite to the
  displacement of the other wave.
• y1(x,t) - y2(x,t)
• From linear superposition we find
• Yt(x,t) 0

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What is the distance between successive nodes?
?x
x1, n
x2, n1

• Let a node occur at a point x1 and have integer
  value n
• The neighbouring node occurs at x2 and has
  integer value n1.
• kx1 np
• kx2 (n1)p
• k(x2 - x1) k?x p
• But k 2p/l
• 2p?x/l p
• ?x l/2

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Key Features of Standing Waves

• There are points on the string where the
  displacement of the standing wave is a maximum.
• These points are known as ANTI-NODES and occur
  spatially when
• sin(kx) 1
• This is satisfied when
• kx (n1/2)p (n an integer 0,1,2..)
• To see these point the time must satisfy wt mp
  with m an integer.

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What is the distance between successive
anti-nodes?
x2, n1
?x
x1, n

• Let an anti-node occur at a point x1 and have
  integer value n
• The next anti-node occurs at x2 and has integer
  value n1.
• kx1 (n1/2)p
• kx2 (n11/2)p
• k(x2 - x1) k?x p
• But k 2p/l
• 2p?x/l p
• ?x l/2

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Example Separation between adjacent nodes
A string of mass per unit length 2 g/m is placed
under a tension of 20 N. Two counter propagating
harmonic waves that vibrate at the same frequency
of 1000 Hz are excited on the string and a
standing wave is formed. What is the spacing
between successive nodes?
We know that ?x l/2 with l the wavelength. And
l v/f with v the wave speed and f the
frequency. But
?x
with m the mass per unit length and F the tension
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So the wavelength is given by
Hence the spacing between adjacent nodes is
As µ 2 x 10-3 kgm-1, F 20 N and f 1000
Hz ?x 5 x 10 -3 m.