Standing Waves and Resonance Standing Wave: “Standing waves” are formed from two or more...

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Standing Waves and Resonance

Standing Wave: “Standing waves” are formed from two or more traveling waves that collide and are “in tune” with one another in such a way that their amplitudes add or subtract in repetitive ways. This is called “Resonance”.

Wave moving right

Wave moving left

Standing Waves and ResonanceGiven a string whose length is L:

The longest standing wave possible is called the “Fundamental”, or “1st harmonic”.

L

Standing Waves and Resonance

What fraction of a sine wave is this?

L

Standing Waves and Resonance

L

What fraction of a sine wave is this? ½ wave

Therefore, using “ L”, how long is one full sine wave (l)

Standing Waves and Resonance

What fraction of a sine wave is this? ½ wave

Therefore, using “ L”, how long is one full sine wave (l) = 2L

L

Standing Waves and Resonance

1st harmonic

2nd harmonic

3rd harmonic

4th harmonic

Standing Waves and Resonance

1st Harmonic:

2nd Harmonic:

3rd Harmonic:

4th Harmonic:

L

l = 2L

l = ___?

l = ___?

l = ___?

Standing Waves and Resonance

1st Harmonic:

2nd Harmonic:

3rd Harmonic:

4th Harmonic:

L

l = 2L

l = ___?

l = ___?

l = L

Standing Waves and Resonance

1st Harmonic:

2nd Harmonic:

3rd Harmonic:

4th Harmonic:

L

l = 2L

l = ___?

l = L

l = 2L/3

Standing Waves and Resonance

1st Harmonic:

2nd Harmonic:

3rd Harmonic:

4th Harmonic:

L

l = 2L

l = L

l = 2L/3

l = L/2

Standing Waves and Resonance

What’s the pattern here?

1

22

LL

2

2LL

3

232 LL

4

2

2

LL

n

L2

,...4,3,2,1n

In general,

where:

Resonant wavelength formula

Standing Waves and ResonanceNomenclature:

Node

Antinode

Node: a region of zero amplitude

Antinode: a region of maximum amplitude

Standing Waves and ResonanceResonant Frequencies Of Harmonic Standing

Waves:

We learned l = 2L/n since v = fl,

n

Lfv

2

Standing Waves and Resonance

)2( Lfnv

n

Lfv

2

Resonant Frequencies Of Harmonic Standing Waves:

We learned l = 2L/n since v = fl,

Standing Waves and Resonance

fvL

n

2

n

Lfv

2

)2( Lfnv

Resonant Frequencies Of Harmonic Standing Waves:

We learned l = 2L/n since v = fl,

Standing Waves and Resonance

fvL

n

2

n

Lfv

2

)2( Lfnv

vL

nfn

2,...4,3,2,1nwhere:

Resonant frequency formula

Resonant Frequencies Of Harmonic Standing Waves:

We learned l = 2L/n since v = fl,

Standing Waves and ResonanceTherefore the 1st resonant frequency

corresponds to the 1st resonant wavelength…

--- 1st Harmonic

--- 2nd Harmonic

--- 3rd Harmonic

--- 4th Harmonic

1

21

L

2

22

L

3

23

L

4

24

L

vL

f

2

11

vL

f

2

22

vL

f

2

33

vL

f

2

44

Standing Waves and Resonance

Standing Waves and Resonance

What two physical characteristics play a role in determining the velocity on a string?

Standing Waves and Resonance

What two physical characteristics play a role in determining the velocity on a string?

1. Tension

2. Mass, or inertia

Standing Waves and Resonance

What two physical characteristics play a role in determining the velocity on a string?

1. Tension

2. Mass, or inertia

What do we expect the relationship with velocity to be?

Standing Waves and Resonance

What two physical characteristics play a role in determining the velocity on a string?

1. Tension

2. Mass, or inertia

We expect the relationship to be as follows:

massv

Fv

1

Standing Waves and Resonance

The mass is applied by using density; specifically linear density:

length

masslinear density

surface density

volume density

area

mass

volume

mass

Standing Waves and Resonance

Finally, the equation used to find velocity can be described:

F

v F = tension

m = linear density

Units = m/s

Standing Waves and Resonance

Wave Velocity on string:

F

v

where F = Forcestring oflength

string of mass

Problem: A force of 45 Newtons is applied to a string with a mass of 0.012 kg and a length of 1.5 meters. Find the velocity of a wave on that string:

Standing Waves and ResonanceSound Velocity on string:

F

v

where F = Force , and

string oflength

string of mass

Example: A force of 45 Newtons is applied to a string with a mass of 0.012 kg and a length of 1.5 meters. Find the velocity of a wave on that string:

m/s 75/5625008.0

4545

5.1012.0

kgNmNNF

vmkg