A sinusoidal motion is one in which each part of the object is in simple harmonic motion at the same frequency

These motions are called NORMAL or NATURAL MODES Every possible free vibration

of an object is a sum of its natural modes.

What about n>2 coupled pendulums?

First, how many natural modes are there?

For two coupled pendulums

(swinging in the plane of the page)

there were two modes.

For two un-coupled pendulums,

(swinging in the plane of the page)

there are…

Two modes ! (What are they?)

What about n>2 coupled pendulums?

For n pendulums

(swinging in the plane of the page) there are n natural modes.

•If they are uncoupled, we may take each natural mode to be the swinging of one pendulum.

•If they are coupled, the modes are complex; each mode will involve motion of more than one pendulum.

Note that it is NOT true that in coupled modes all pendulums must move… look at Mode 2!

A point that does not move is called a NODE of the mode.

But how can I tell what the natural modes of n coupled pendulums look like?

We’d like to be able to draw the pendulums when they at the maximum displacement, for each mode, as done at right (from your book). Here’s a graphical method that works…

1. Draw an nxn grid, where n= number of pendulums.

2. Treat each horizontal line as the x-axis on a graph.

3. Sketch a sine wave on

each, starting from 0 waves, increasing by 1/2 wave each line. This takes some practice to get the waves nice. Pay attention to where the zero-crossing(s) go! The ends are always maxima (±) !

4. You can now read off the displacement for each pendulum (P1- P5) on the

y-axis of each sine wave. Pay attention to signs.

Mode 1

Mode 2

Mode 3

Mode 4

Mode 5

Mode 1

Mode 2

Mode 3

Mode 4

Mode 5

POSITION

Left of

Right of

Center Center

Mode 1

Mode 2

Mode 3

Mode 4

Mode 5

In mode 3, which pendulums DO NOT MOVE AT ALL ?

A] P3, at the center

B] P2 & P4

C] P1 & P5

D] all move

E] none move

Note that the number of nodes in the sine wave = mode# + 1

Mode 3 has nodes at P2 & P4

OK, now I can draw modes. How do I know which has the highest and which has the lowest frequency?

1. Higher modes have higher frequencies

2. The more the coupling springs are involved, the higher the frequency.

Consider the coupling springs to be in one of five states (here).-Highly compressed - score 2-Slightly compressed - score 1-Relaxed - score 0-Slightly stretched - score 1-Highly stretched - score 2

Figure out how much the forces from the adjacent springs are pushing on a mass.

Consider the coupling springs to be in one of five states (here).-Highly compressed - score 2-Slightly compressed - score 1-Relaxed - score 0-Slightly stretched - score 1-Highly stretched - score 2

Figure out how much the forces from the adjacent springs are pushing on a mass.

This mass is being pushed to the right by a slightly compressed left spring, and pulled to the right by a slightly stretch right spring. Score = 1+1=2.

This mass is being pushed the right by a highly compressed left spring, and pulled to the right by a highly stretched right spring. Score = 4.

Note: ignore ends and masses that don’t move.

Mode 1

Mode 2

Mode 3

Mode 4

Mode 5

Higher modes always involve more contribution from the coupling springs, because in higher modes, the neighboring pendulums are doing different things.

So higher modes have higher frequencies!

I really don’t know what the frequencies of these modes are…

but I will guess.

Consider a mass that moves. It moves because of the restoring force of gravity, plus a spring contribution.

Contribution from coupling springs:

0

1

2

3

4

€

f =1

2π

k

mSpring & Mass alone, F=-kx

€

fn =1

2π

(n −1)k + mγ

m

Restoring force in mode n

m is the restoring force of gravity, it turns out that = g/L

In real life, pendulums can move both in the plane of the page, and out of the plane. We say that each pendulum has 2 “degrees of freedom.”

A single pendulum therefore has TWO perpendicular MODES. They have the same frequency.

(We physicists like to use the word “degenerate” to describe when multiple modes have the same frequency. I’m not sure where the terminology comes from… use your imagination?)

So five coupled pendulums have 10 modes. The other 5 are (looking down from above):

Any free vibration of a system with N degrees of freedom is just a sum of the N natural modes, each added in with a possibly different amplitude (& phase)

The sine wave construction works for these modes too!

We are halfway to understanding ALL musical instruments.

If I give you an object, you just need to find the natural modes of vibration. Those are the frequencies that may go into the waveform (timbre) of the instrument.

The second part is: how do I know how much of each natural mode (overtone) to include in my waveform? We’ll address that in short order.

But first… vibrations of “real” musical objects - metal or wood bars

Solid materials consist of atoms connected by “springs”.

Thus, like the coupled pendulums, each atom has a restoring force that depends on the positions of its neighbors.

For solid bars, the finite thickness of the bars adds some complications, and the sine wave construction (for drawing the modes) doesn’t work perfectly.

2nd mode. (NOT second harmonic… it’s not “harmonic”)

Tuning Fork

Huh? Combining sine waves that are NOT in a harmonic series should NOT give a tone of “definite pitch”.

Tuning Fork

Remember: the natural modes only tell you what sounds (waveforms) are allowed. ANY SOUND THAT IS NOT A SUM OF THE NATURAL MODES is NOT allowed.

I can tell you from this graph that a tuning fork CANNOT SOUND like a bassoon.

Tuning Fork

Certainly, a pure vibration in mode 1 IS allowed.

That would give a pure sine wave, with a pure pitch.

In order to understand what instruments sound like (not just what they can’t sound like) we need to understand

1) Which modes get “excited”

2) How rapidly modes decay

Striking an object at any point excites each natural mode in proportion to how much that mode involves motion of the struck point.

If I strike the bar about 3/4th of the way out, as shown by the arrows, which of the modes shown are “excited”?

A] first mode f1

B] second mode f2

C] third mode f3

D] f1 & f3

E] all threef2 is NOT excited. That mode involves NO motion at the striking point.

When an object is struck at a node of a mode, that mode DOES NOT CONTRIBUTE to the motion.

We’ll call the amplitudes of the modes that do contribute the “vibration recipe” (just as we talked about a recipe for a particular sound.)

So we can get our tuning fork to omit f2. But it will still have a lot of f3, f4, f5 etc, which are anharmonic overtones.

There are two (more) reasons why a tuning fork works to generate a “pure” pitch.

1. Modes decay at different rates. For a tuning fork, the higher modes decay quickly, leaving the pure fundamental tone. (Note: it is NOT universally true that higher modes decay faster than lower modes.)

2. In actual use, we place the base of the tuning fork on a resonant surface. The higher modes don’t move the base of the fork much, so we don’t hear them! Spectrum of a tuning fork - demo

How does a spectrum analyzer work? -Answer: there is a mathematical procedure to extract the amount of each sine wave in a complex waveform.

Sound (pressure waves) from a tuning fork

Immediately after striking

A few seconds after striking

Let’s make a guess as to the main frequency contributing to these “spikes”

The frequency required to make a “spike” like this looks to beA] about half the fundamental, f1/2B] about twice the fundamental, 2f1

C] less than one fifth of the fundamentalD] more than five times the fundamental

Sound (pressure waves) from a tuning fork

Immediately after striking

The spikes require a frequency more than 5 times the fundamental. In fact, the spikes are caused by the first overtone (the fork was not struck at the node)

“Clang tone”

Using mathematics (differential equations), we can describe the forces between different parts of an object and, in principle, find the natural mode frequencies.

For most objects, the math is so hard, it can only be done on a computer.

But there is a way to find the natural modes (at least some of them) by experiment. How?

Gong show.

The idea of “hitting” something and then “listening” to it ring down is not limited to acoustics.

We can use an radiowave “hammer” to hit all the hydrogen nuclei in a molecule. The nuclei vibrate, and, because they are charged, emit radiowaves. A spectrum analysis of the radiowaves gives:

Just as with our acoustic experiment, different objects (molecules) will give different spectra.

Maybe we could hit YOUR BRAIN with our radiowave hammer…

Magnetic Resonance Imaging is simply “listening” to the radiowaves emitted by hydrogen nuclei in the body, after hitting them with a radiowave hammer.

Back to music!

How many nodes are there for the 6th mode?

A] 5B] 6C] 7D] 12

It is always true that increasing the mode number by 1 increases the number of nodes by 1, for an extended object

A uniform bar, unclamped, in vibration

This bar will only give a clear pitch if, like the tuning fork, the higher modes decay away very quickly.

If so, we will have a musical instrument that sounds like a tuning fork.

Is there another way to make this musical?

A uniform bar, unclamped, in vibration

Here, the bending stiffness plays the role of a spring constant. If you made bars that were easier to bend (keeping the mass the same) the mode frequencies would:

A] get higherB] stay the sameC] get lower

In the middle of the bar, only the ODD MODES are bending.

We can change the stiffness of the bar IN ITS MIDDLE SECTION by thinning it.

Of course, thinner bars are easier to bend than thicker ones.

Middle-Thinned Bar Uniform Bar

Ignoring the change in mass, the bar above should have mostly:

A] even modes with lower frequencies than those of the uniform bar

B] odd modes with lower frequencies than those of the uniform bar

Middle-Thinned Bar Uniform Bar

f1’ = 0.69 f1

f2’ = f2 = 2.76 f1Now, the ratio of f2’ to

f1’ is 4:1 !That gives a two-octave overtone, and a pleasing pitch.

We don’t need a complete harmonic series to sense a pitch, just a couple of the low harmonics. (Perhaps our brains seek order and beauty in chaos!)

Middle-Thinned Bar

f1’ = 0.92 f1

f2’ = f2 = 2.76 f1We could have thinned to lower f1’

only to 0.92 f1. That would give a ratio f2’: f1’ = 3, also a small integer. What note in the harmonic series would that give?

A] an octave above f1

B] an octave and a major third above f1

C] an octave and a fifth above f1

f1’ = 0.92 f1

f2’ = f2 = 2.76 f1

Each thinned bar shifts the 1st mode frequency down. Which bar shifts it down more?

f1’ = 0.69 f1

f2’ = f2 = 2.76 f1

A

B

Xylophone

Marimba

Drumheads and the Quantum Mechanics of Atoms

Using differential equations, we can carry this model of a drum to the “continuum limit”, where each mass is infinitesimal.

The membrane tension plays the role of the restoring force.

•These show whether the drum membrane is bulging out (+) or in (-) at a given instant in the vibration. •A half cycle later, all the + change to - and vice versa.•We have nodal lines, rather than nodal points.

Note: vertical & horizontal are not special. (The drum is symmetric.) Mode 2, for example, is degenerate with its 90° rotation; Mode 3 with its 45° rotation.

http://www.falstad.com/circosc/

What mode is this? A] f1 B] f2 C] f4 D] f6 E] f9

Mode 1 2 3 5 7 10

Recall our discussion of vibration recipes, but think about degeneracy.

If you strike a drum at the point shown at left, which of the modes shown (or their degenerate partners) can you excite?

A] none of themB] only mode 1C] all of them

Let’s try the spectrum analyzer!

All things are waves, including electrons. The electron wave is trapped in the atom, just like the drumhead is attached to the drum. Do these electron waves remind you of drum modes?

How about this electron wave?(Electron waves in atoms are called “orbitals”)

We shouldn’t get carried away - there is not an exact mapping of drumhead modes onto orbitals… the former are 2D vibrations and the latter 3D.

Drums generally don’t give us a good sense of pitch. That makes sense, given the anharmonic overtones:

1.592.142.302.652.923.163.503.603.65

But what about…

Timpani (singular - Timpano)

First, a little more about “vibration recipes.”Recall that:

Striking an object at a point excites each natural mode in proportion to how much that mode involves motion of the struck point.

We add:

If you strike several points at once (or a region), you get the same recipe as adding together the recipes for striking at each point individually.

Be careful! You need to pay attention to signs. If you strike a + and a - region at the same time, you can’t excite that mode.

Of course, + and - switch every half cycle, so we have a corollary:

a striking force with duration T can only excite modes with periods > 2T,

i.e. frequencies < 1/(2T).

Mode 1 2 3 5

If you strike a drumhead simultaneously at the two points shown at left, what modes (including their degenerate partners) above can you excite?

A] none of themB] only mode 1C] only modes 1 and 3D] modes 1,3, and 5E] all of them

The timpani mallet is usually a few centimeters across. To figure out which modes are strongly excited, you have to see whether the points struck by the mallet are all “in phase”. When striking about halfway out, modes with circular nodal lines are only very weakly excited.

Two other facts account for the pitch of a timpano.

1) Mode 1 decays very quickly. (It radiates sound very well, because the whole membrane moves up & down. So it makes a loud but brief BOOM.)

2) The bowl changes the frequencies of the other modes.

In a kettle drum, the motion of the membrane requires air in the kettle to move too. That increases the effective mass (think mass & spring)… so it will

A] lowerB] raise

the frequencies of the modes.

The frequencies are in approximate ratios of 2:3:4:5:6

The pitch you “hear” on a timpano is not actually there… it’s the octave BELOW the lowest long-lasting mode.

All objects have natural modes of vibration Handbell

Pressure above ambient (N/m2)

Vibrations decay in time because of• friction• radiation (of sound)

The amplitude decays by half for each time interval t1/2

How many dB does the sound intensity level fall in each t1/2?A] 0.5 dB B] 3 dB C] 6 dB

As we have seen, the changing overtones change the timbre of the sound. This sound changes over time.

The “damping time” is the time it takes the intensity to fall by a factor of a million (106). How many dB is this?

Ans 60 dB. Since 6 dB gives

t1/2, is 10 amplitude half-times.

One final comment about “vibration recipes.”

We know if you strike an object, you don’t excite any modes that have nodes at the striking point.

You can add to the friction of a vibrating object by touching it. That kills any mode that wants to make the touched point move.

If a free bar is vibrating in all modes, and you touch it in the middle, what mode(s) are “killed”?

A] All of themB] All even modesC] All odd modesD] Only mode 1 is killed

Vibrating StringsStanding Waves = A left-going wave + A right-going wavehttp://www.walter-fendt.de/ph14e/stwaverefl.htm

•Waves that travel down a string with speed v = are allowed “excitations”.

T=tension, mass per unit length.

•If the ends of the string are fixed, we need to consider only those excitations that fulfill the “boundary conditions”, i.e. that leave the ends stationary.

•A “standing wave” works if it has the right wavelength!

€

T

μ

These are the first four modes of a vibrating string. What are their mode frequencies?

V=f

We just need their wavelengths! Each “loop” is half a wavelength, and mode n has n loops, so

€

T

μ

€

n ⋅λ n2= L

λ n =2L

n

These are the first four modes of a vibrating string. What are their mode frequencies?

V

€

= fnλ n = fn2L

n=T

μ

€

fn =n

2L

T

μ

Note that the overtones are perfectly harmonic (integer multiples of f1 ) - demo

Typical guitar: T= 150 N, = 0.005 kg/m, v= 170 m/sPiano T=650 N, v=330 m/s. Higher wave speed gives higher freq for same L

€

fn =n

2L

T

μ

What is the effect of doubling the length of a string on the pitch of the fundamental? (Keeping tension & mass constant.)

A] lowers it an octaveB] lowers it a fifthC] raises it a fifthD] raises it an octave

€

fn =n

2L

T

μ

If you want to raise the pitch of the fundamental by an octave, what do you do to the tension (keeping L & mass constant)?

A] reduce tension by halfB] double tensionC] quadruple tension

Vibration Recipes for Plucked or Hammered Strings

Take a stab at this:If you pluck a string exactly at its midpoint, what modes are absent from the subsequent motion?

A] all odd modesB] all even modesC] all absent except fundamentalD] none are absent

Mode 1

Mode 2

Mode 3

Mode 4

Another plucking result from Fourier analysis: aside from modes that are missing/weakened because of plucking at/near a node, the overall ‘envelope’ of the spectrum of modes drops 6 dB/octave (energy per mode) - dashed line

If a plucked string has the mode spectrum shown, how far from the end was it plucked?A] L/5 B] L/10 C] could have been either

Mode 1

Mode 2

Mode 3

Mode 4

Just because an object has a particular distribution of energy in its various modes, does NOT mean that the sound from it will have the same distribution of energy in the overtones.

We have to consider how the motion of the vibration is converted into sound.

Easiest case: electric guitar. “Pickups” measure the motion of the string where they are positioned.