WavesPart 3A: Standing Waves

Last modified: 24/01/2018

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SuperpositionStanding Waves

DefinitionNodesAnti-NodesStanding Waves Summary

Standing Waves on a StringStanding Waves in an Open-Ended PipeStanding Waves in a Pipe with One Closed End

Superposition Contents

If two (or more) waves are travelling in the same medium, in the samedirection, then the two displacements y1 and y2 will add together:

y1 + y2 = F (x − vt) + G(x − vt) = H(x − vt)

Recall that a travelling wave has the form F (x − vt) and the speed v willbe the same for all waves in the same medium.

The sum, or superposition of two (or more) waves travelling inthe same direction is also a wave, in htat same direction.

The situation is a little different for waves travelling in differentdirections.

Standing Waves Contents

Standing (or Stationary) waves are formed by the superposition ofwaves travelling in opposite directions, in particular by waves that areotherwise identical. This will most often occur when a wave meets itsown reflection.In this case, if y1 = F (x − vt) and y2 = F (x + vt), then the superposition,

y1 + y2 = F (x − vt) + F (x + vt)

will not be a function of x − vt and so is not a travelling wave.

As usual, let’s consider only harmonic waves, with y1 = A sin(kx − ωt)and y2 = A sin(kx + ωt). Then the superposition will be:

y1 + y2 = A sin(kx − ωt) + A sin(kx + ωt)= 2A sin(kx) cos(ωt)[

Using the identity: sin A + sin B = 2 sin(A+B

2)

cos(A−B

2)]

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The superposition of the waves has the following form:

y(x , t) = 2A sin(kx) cos(ωt)

As already mentioned, this is clearly not a travelling wave. There ishowever, still some vibration occuring. To understand what is going on,let’s plot this equation over half a period T :

t=0t = 1

12 Tt = 1

6 Tt = 1

4 Tt = 1

3 Tt = 5

12 Tt = 1

2 T

Unlike a travelling wave, where every position vibrates with the sameamplitude, in a standing wave the amplitude of the vibration varies withposition x , and at some particular values, this amplitude is zero.

Nodes Contents

Examining the equation, we see that the minimum amplitude occurswhen sin(kx) = 0.

Remembering that k = 2πλ where λ is the wavelength of the original wave:

sin(kx) = 0⇒ kx = nπ n = 0, 1, 2 . . .

x = nπk

= n�πλ2�π

= n2λ

At these values of x the amplitude of the vibration will always be zero.

These points are called the nodes of the vibration.

The distance between two adjacent nodes is 12λ.

Anti-Nodes Contents

The maximum vibration amplitude occurs when sin(kx) = ±1.

sin(kx) = ±1⇒ kx = (n + 12 )π n = 0, 1, 2 . . .

x =(n + 1

2 )πk

=(n + 1

2 )�πλ2�π

= (2n + 1)4 λ

At these values of x the vibration will have a varying displacement(because of the cos(ωt) factor), but the largest amplitude. These pointsare called the anti-nodes of the vibration, and are located half-waybetween the nodes.

Note that as in a travelling wave, the individual points of themedium are vibrating around their equilibrium position. BUT themaxima and minima of the vibration are at fixed points. The waveis not travelling, it is stationary.

Standing Waves Summary Contents

Standing waves consist of vibrations in a medium which have fixed pointsof minimum (i.e. zero) vibration (the nodes) and maximum vibration(the anti-nodes). In what follows, we will indicate a standing wave byshowing the range of maximum vibration:

nodes

anti-nodesλ2

λ2

The distance between two adjacent nodes, or two adjacent anti-nodes isthe same: λ

2 , where λ is the wavelength of the original harmonic wavecreating the standing wave.

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A standing wave is represented by the function:

y(x , t) = 0.004 sin(100x) cos(600t)

Find the amplitude, frequency, wavelength and speed of the compo-nent travelling waves. Also determine the distance between nodesin this standing wave,

Comparing the general and given equations gives:

y = 2A sin(kx) cos(ωt)y = 0.004 sin(100x) cos(600t)

So: 2A = 0.004⇒ A = 0.002 m = 2 mm

ω = 600⇒ f = ω

2π = 6002π = 96 Hz

k = 100⇒ λ = 2πk = 2π

100 = 0.063 m = 6.3 cm

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To find the speed v :

v = f λ = 96× 0.063 = 6.05 m/s

And the distance between nodes:

12λ = 1

2 (0.063) = 0.032 m = 3.2 cm

Standing Waves on a String Contents

Many musical instruments (e.g. guitar, violin, piano etc) use a length ofstring (or wire) fixed at both ends. Sound waves are generated bystanding waves on these strings.

The fact that the ends of the strings are fixed means that there must benodes located at the end-points.

This restriction means that only a limited number of standing waves arepossible on the string, and thus a limited number of sound frequencies(i.e. musical notes) can be produced . The possible standing waves arecalled the modes of vibration of the string.

Note that this situation will apply to other objects also. All objects willhave a limited set of possible vibrational modes. In general these modeswill be much more complicated (occuring in 2 or 3 dimensions) thanthose we are about to see for the string.

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The simplest mode of vibration for a string is that with the least numberof nodes - only those at the two ends of the string.

1st Mode

L = λ12 = v

2f1

⇒f1 = v2L

L

The next simplest has three nodes:

2nd Mode

L = 2× λ22 = v

f2

⇒f2 = 2( v

2L

)= 2f1

L

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3rd Mode

L = 3× λ32 = 3v

2f3

⇒f3 = 3( v

2L

)= 3f1

4th Mode

L = 4× λ42 = 2v

f4

⇒f4 = 4( v

2L

)= 4f1

The pattern should be clear. For any integer n we have for the n-th mode:

L = n × λn2 = nv

2fn⇒ fn = n

( v2L

)= nf1

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These vibrational frequencies are the musical notes that the string cancreate.

The lowest frequency, f1 is the fundamental frequency.

In cases like the string where the frequencies of higher modes are integermultiples of the fundamental frequency, then these allowed frequenciesare called harmonics.

For a stretched string, the frequency of the n-th mode (aka then-th harmonic) is fn, given by;

fn = nf1 = n( v

2L

)= n

2L

√Tµ

where we remember that for a wave on a string v =√

Tµ with T

being the tension, and µ the linear mass density, of the string.

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A mass M = 5 kg is suspended by a string with linear mass densityµ = 0.40 g/m over a frictionless pulley as shown below right.

Calculate the fundamental frequencythat the string will vibrate with.Find also the frequencies of the secondand third vibrational frequencies. M

L = 20 cm

The block M is in equilibrium, so T = Mg = 50 NFor the fundamental mode:

f1 = v2L = 1

2L

√Tµ

= 12(0.2)

√50

0.4× 10−3 = 884 Hz

For the second mode: f2 = 2f1 = 2× 884 = 1770 Hz

And the third mode: f3 = 3f1 = 3× 884 = 2650 Hz

Standing Waves in an Open-Ended Pipe Contents

Other musical instuments (e.g. flutes, organs) produce sound viastanding waves in an open-ended pipe, formed by the superposition ofsound waves. Again there are physical restrictions on these allowedmodes - in this case, each open end of the pipe must be an anti-node.

The simplest mode of vibration for the pipe is that with the least numberof nodes - only one at the middle of the pipe..

1st Mode

L = λ12 = v

2f1

⇒f1 = v2L

L

The speed v here is the speed of sound in the air inside the pipe.

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2nd Mode

L = 2× λ22 = v

f2

⇒f2 = 2( v

2L

)= 2f1

L

3rd Mode

L = 3× λ32 = 3v

2f3

⇒f3 = 3( v

2L

)= 3f1

4th Mode

L = 4× λ42 = 2v

f4

⇒f4 = 4( v

2L

)= 4f1

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These equations should look familiar!

The expression for the frequencies of the vibrational modes (har-monics) of the open pipe is the same as for the stretched string:

fn = nf1 = n( v

2L

)In practice, this formula is not quite accurate. Because of the inertia ofair at the ends of the pipe, the above formula must be slightly modifiedby what is called the end correction, involving the radius r of the pipe.

fn = n( v

2L′)

where L′ = L + 1.2r

This is an empirical formula, i.e. derived from experiment rather thanmathematically.

Pipe with One Closed End Contents

Another musical possibility is a pipe with one end open and the otherclosed. As we should expect, the possible vibrational modes are restrictedby the conditions that the closed end is a node, and the open ananti-node.

The simplest mode of vibration includes only the required node andanti-node:

1st Mode

L = 12 ×

λ12 = v

4f1

⇒f1 = v4L

L

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2nd Mode

L = 32 ×

λ22 = 3v

4f2

⇒f2 = 3( v

4L

)= 3f1

L

3rd Mode

L = 52 ×

λ32 = 5v

4f3

⇒f3 = 5( v

4L

)= 5f1

4th Mode

L = 72 ×

λ42 = 7v

4f4

⇒f4 = 7( v

4L

)= 7f1

Contents

The pattern this time is a little trickier:

fn = (2n − 1)( v

4L

)= (2n − 1)f1 n = 1, 2, 3 . . .

The frequencies of the higher modes are integer multiples of thefundamental fequency, so are harmonics, but this time only the oddharmonics are present, which makes the naming potentially a littleconfusing:

I the second mode is the third harmonicI the third mode is the fifth harmonicI and so on . . .

For the same reasons as with the open ended pipe an empirical endcorrection factor is required in this formula:

fn = (2n − 1)( v

4L′′)

where the corrected length is L′′ = L + 0.6r .