2: Oscillations(Chapter 15)

Phys130, A01/EA01Dr. Robert MacDonald

Oscillating SystemsAll kinds of things oscillate:

• Pendulums

• Buildings

• Machinery

• Circuits

You can describe it all with the same physical and mathematical language. If we can understand a simple system, like a mass on a spring, we have a handle on everything else.

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Demo: Masses & Springs

PhET: Online simulated physics demos. Very useful. Designed especially for you to play with — try this at home!

This demo: “Masses and Springs”, representing our basic model for periodic motion.

You’ll explore this in Lab 2 as well, including a “virtual lab” at the UGL website.

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http://phet.colorado.edu/simulations/sims.php?sim=Masses_and_Springs

Choosing CoordinatesRemember that you can always choose your coordinate system to be anything you want. The right choice of coordinates can make your life a whole lot easier.

This is one-dimensional motion. Let’s call it x, and let’s set up to be positive.

When we hang the mass, the spring stretches a bit. Once it stops bouncing, the spring is barely holding up the mass; this is the equilibrium position. Let’s make x=0 here. Everything will be measured relative to this point.

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50 g

x=0

xFs

Fg

Fs –Fg=

Unstretchedposition

at equilibrium:

Which graph describes the motion best?

Time

Displacement

Time

Displacement

Time

Displacement

Time

Velocity

Time

Accelleration

Describing OscillationsAmplitude (A [m or cm]): The maximum magnitude of the displacement from equilibrium (maximum |x|).

Cycle: A complete round trip (e.g. top-bottom-top).

Period (T [s]): The time for one cycle (“seconds per cycle”).

Frequency (f [Hz or s-1]): The number of cycles in a unit of time (“cycles per second”).

Angular Frequency (ω [rad/s or s–1]): 2π times the frequency ( ω=2πf ). (Yes, it’s weird.)

7

f = 1T

peak

trough

equilibrium position

Time

Displacement

+A

-AT = 1/f

Describing Oscillations

Angular Frequency

This looks exactly like a cosine function, so it would be nice to be able to use that function to describe the motion. But cos takes an angle, and this is a function of time.

Can we come up with a way to shoehorn something to do with time into the cos function and wind up with this graph?

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Time

Displacement

Displacement vs Time graph:

Angular frequency has the units “radians per second”. So if we multiply by time (ωt) we get units of radians — just what we need for the cosine function.

Make sure this gives the right period:! (2πf)T = (2π/T)T = 2π = one complete cycle.So we get a cycle’s worth of radians in one period, which is what we need. This should work!

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cycles per second

Radians per cycle

ω = 2π f

Angular frequency

(not a “w”)

Now consider the amplitude. Cosine has an amplitude of 1 (it ranges between +1 and –1). Our motion has an amplitude of A, so multiply the cos function by A.

Putting it all together, we can say:

x = A cos(ωt)

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Time

Displacement

+A

-AT = 1/f

(ω = 2π f)

Motion of a simple harmonic oscillator.

A mass is attached to a spring, pulled 5 cm from the equilibrium position, and released. It oscillates with a period of 3 s. Write down its equation of motion.

We need to put actual values into the x = A cos(ωt) equation.

From the question, A = 5 cm.

The period T = 3 s. So ω = 2πf = 2π/T = 2π/(3 s). We could leave it like that or write ω = (0.667)π s–1, or ω = 2.094 s–1. Any of these forms would be fine.

Then we have: x = (5 cm) cos[ (2.094 s–1) t] .12

eg: A specific oscillator

• what an oscillator is.

• what its motion looks like on a graph.

• how to describe the motion mathematically.

• relationships between aspects of its motion.

• plenty of vocabulary!

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So now you know: Why does it do that?We can now describe the motion of a simple oscillator in detail. But why does it do that? Why a cosine wave and not, say, a triangle wave? Let’s find out.

Forces are what cause things to move (or not move).

• If there’s no force, an object will either sit still or move in a straight line at constant velocity.

• If it’s not doing one of those two things, there are forces involved.

Knowing what the forces are and how they change will tell us everything about how the object will move.

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When the net force on a system pushes the system back towards equilibrium, we call this a restoring force.

Restoring Force

50 gFs

FgFs Fg<equilibrium

position

Fnet is down50 g

Fs

FgFs Fg>

Fnet is upHere Fnet always restores equilibrium.

So Fnet is a restoring force.

Fnet = Fs –!Fg

For a hanging mass:

Springs

If you stretch or compress a spring, the spring will push back, to try to return to its unstretched length.

How hard the spring pushes back depends on how much you’ve stretched it.

This is Hooke’s Law.

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50 g

50 g

50 g

unstretched position

x

xF

F

Hooke’s LawNamed for Robert Hooke, 17th century British physicist.

He first published it as a Latin anagram, “ceiiinossssttuu”, in 1676. He published the solution in 1678: “Ut tensio, sic vis,” or “As the extension, so the force.”

In modern terms, Hooke’s Law says that F = –kx.

• k is known as the “spring constant” and describes the “stiffness” of the spring.

• The –!sign says the force opposes the displacement.

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When a spring is at its unstretched length, the force it applies is zero.

If you have some object attached to the spring, and you pull it and let go, the spring will pull it back to the unstretched position...

...but when it gets there, the net force is zero and there’s nothing to stop it. So the object sails on through to the other side.

Now the spring is pushing again (the other direction, still back to the unstretched position), and the motion repeats.

The result is periodic motion or oscillation.

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Springs and Restoring ForceSo when the only thing doing any pushing or pulling in a system is a spring, and the spring always tries to move the object back to the equilibrium position, the restoring force is just the spring force, described byF = –kx.

What happens when we hang the mass vertically? What effect does gravity have on the system’s oscillation?

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50 g

50 g

50 g

unstretched position

x

xF

F

equilibrium position (sum of forces = 0) Vertical Restoring Force

50 gFs

Fg

equilibriumposition

Fg = mg (m = 50 grams)

Fs = k(stretch)Fs = k(ΔL-x)

unstretchedposition

xΔL

ΔL–x =

L

Fnet = Fs – Fg

Fnet = k(ΔL–x) – mg

be careful with the signs!

original spring length without the mass

amount the spring is stretched when you hang the mass at rest.

stretch

watch the signs!

?

(from our coordinates discussion)

So what’s ΔL?

mg is the reason the spring stretches. It determines ΔL!

When you hang a mass on a spring, and the whole thing is stationary, how much does the spring stretch?

Fs = Fg at equilibrium (and only at equilibrium!).

k(stretch) = mg

But at rest, stretch = ΔL, by definition.

So k ΔL = mg.

Then Fnet = mg - kx - mg, or: Fnet = –kx.Restoring forcefor a mass on a spring.

Fnet"= k(ΔL–x) – mg" = kΔL –!kx – mg

It’s the same as for a horizontal spring! Neat!

Restoring Force in GeneralIn any situation where the net force always points back towards some “equilibrium” position, the net force is a restoring force.

• For a horizontal spring, the restoring force is Fs.

• For a vertical spring, the restoring force is Fs – Fg.

• For a pendulum: At rest, it points straight down. If you pull it to one side, gravity will try to make it point straight down again. In this case gravity is the restoring force.

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Simple Harmonic MotionThe simplest kind of oscillation happens when the restoring force F is directly proportional to the displacement x from equilibrium.

Since, for the mass on a spring, F = –kx (assuming an ideal spring), the mass on a spring will move in SHM.

(If it’s not an ideal spring, the motion won’t look like a cosine anymore. It’ll do funny things, especially at the peaks and troughs.)

So why does an object’s motion look like a cosine when Fnet = –kx?

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Describing SHMThe relationship between the forces on an object and how that object moves is Newton’s Second Law, F = ma.

(Accelleration is not constant here!)It turns out the solution to this differential equation is x = A cos(ωt). So that tells us what the motion is!

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Proof Using CalculusIt’s fairly simple to show that x = A cos(ωt) is a good solution to a = –(k/m)x using calculus:

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So it works, as long as ω2 = k/m. We’ve now determined the behaviour of the mass using just Newton’s second law of motion, and the force.

(chain rule)

(chain rule)

a = − k

mx! compare to

So what does it mean?

The mass of the bouncing object (m) and the stiffness of the spring (described by the spring constant k) completely determine the frequency of oscillation.

To check: SI units on m are ___. SI units on k are ___. So the units on ω work out to ____.

(Radians don’t show up, but radians aren’t really a unit.)

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So we found that or

SHM in the Real World

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We’ve been talking about an ideal spring here. A real spring won’t follow F = –kx if you stretch it far enough, and a lot of other oscillating things won’t either. But for small oscillations (small |x|), it’s approximately true, and often good enough!

Ideal case(Hooke’s Law)

Real springs etc often do things like this.

All three are restoring forces:• F < 0 when x > 0• F > 0 when x < 0

Net force, Fnet

Displacement from equilibrium, x

For small displacements,they still follow Hooke’s Law.

So now you know:• how Hooke’s Law describes springs.

• what a restoring force is.

• how the restoring force in a mass & spring system depends on the position of the mass, whether horizontal or vertical.

• what simple harmonic motion (SHM) is.

• why F = –kx tells us that x = A cos(ωt).

• how the frequency of a simple harmonic oscillator depends on the properties of the oscillator.

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We’ve now covered everything you need for your first assignment!

The cosine function isn’t the only trig tool that’s useful here. It turns out we can describe Simple Harmonic Motion using the same language we use to describe the shadow of an object moving in a circle.(Honest, this is useful.)

Aside: Circular Motion

Fig. 15-12

Galileo measured the angle he saw between Jupiter and Callisto over time.

Callisto is moving in a circle.

Galileo’s observations look like SHM...

Shadow Position...

θθ = ωt

(not a “w”)

A x(t) = A cos(θ)

so x(t) = A cos(ωt)

Fig. 15-13a

... & Velocity and Acceleration

θ

θ

vP’ (constant!)

vP = –vP’ sinθvP

θ

Fig. 15-13cFig. 15-13b

aP’ (const

ant!)

aP

aP = –aP’ cosθ

P’ is a particle moving in a circle. P is a projection moving in SHM.

Magnitude of vP’ and aP’

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Remember that θ = s/r (or arc length s = θr).

Angular speed [rad/s]: v = s/t, so ω = v/r (or v = ωr).

• so on the reference circle: vP’ = ωA.

• Recall angular frequency: ω = 2πf. f is the number of cycles per second, and there are 2π radians per cycle. The same ω!

Centripetal (centre-pointing) accelleration: a = v2/r.

Substitute for v to get a = (ωr)2/r = ω2r.

• so on the reference circle: aP’ = ω2A

(Remember: ω = θ/t)

Back to the ShadowsRecall that the acceleration of the shadow point P is given by a = –aP’ cosθSo if aP’ = ω2A, thena = –ω2A cosθ.

But x = A cosθ!

So a = –ω2x...

Compare with SHM:

They’re the same, if ω2 = k/m!

We already know this, of course, but this confirms that this circular motion business can be used to study SHM.

θ

Fig. 15-13c

aP’ (const

ant!)

aP

aP = –aP’ cosθ

What we know about Simple Harmonic Motion

Restoring force in SHM looks like F = –kx.

Motion vs time is sinusoidal.

• x = A cos(ωt) v = –ωA sin(ωt) a = –ω2A cos(ωt)

Frequency depends on the system, not the initial conditions (such as amplitude).

• Specifically:

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Amplitude vs PeriodA bit of explanation on why the amplitude and period are not connected:

• Pull the mass farther away from equilibrium, and when you let go the restoring force is larger (sinceF = –kx). The mass gets pulled back to equilibrium harder, and its average speed is higher overall — which compensates for having farther to go!

This is why tuning forks don’t change pitch when you hit them harder.

(If you have an oscillator whose period does depend on amplitude, it’s not Simple Harmonic Motion.)

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Coordinates for Horizontal SHM

We’ll be working with a horizontal mass-on-a-spring shortly, so let’s define a coordinate system.

Since the mass is moving horizontally, let’s call that axis x. Let’s make right the positive direction, arbitrarily.

Again, we’ll set x = 0 to be the equilibrium position. Since the only force is the spring force (no gravity to worry about), Fnet = 0 where Fs = 0 — which is the spring’s unstretched position.

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50 g

unstretchedposition

x

x = 0

Fnet = Fs = –kx

Example: Diana, Duck of Science!Diana, Duck of Science! wants to investigate simple harmonic motion. In the name of Science! she has attached herself to a spring.

She pulls on the spring, and finds that a force of 6 N stretches the spring by 0.030 m, so k = ____________________.

Now she pulls herself 2 cm (0.02 m) to the right (positive x) and lets go.

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m = 0.500 kg

ice (no friction)

Quark!

a) What is the period of oscillation?

b) Draw a graph, showing three cycles of oscillation.

c) Where is Diana at t = 0.5 s?

d) What is Diana’s velocity at t = 0.5 s?

e) When does Diana first pass through x = –1.0 cm?

f) What is Diana’s maximum displacement?

g) What is Diana’s maximum velocity?38

Time

Displacement(x

)

(t)

Trig functions require an angle, so we invented one.

The phase is the (fictional) angle that identifies a point in the oscillation cycle.

It’s the angle on the reference circle that gives you the cosine you need.

The initial phase is the phase at t!=!0. So it depends entirely on when you start your watch.

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Phase Angle

θA

Fig. 15-13a

phase angle

Finding the Phasecosθ = x/A, or x = A cosθ.

θ here is the phase; it’s just the angle we have to set the turn table at to get the shadow (P) at position x.

But wait! There are two positions we can put the ball (P’) to get the shadow at the same place!

• These are the phases θ and (2π–θ).

The shadow crosses each x position twice per cycle, going to the left (negative velocity) and to the right (positive velocity).

• So we need both x and v to find θ.

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P’

θA

x

Px

ω

light

P’

2π–θ

A–A

Phase in Trig FunctionsJust like with the turntable, we can see on the graph the two phase angles (θ and 2π–θ) that give the same cosine, and hence the same position x. (x = A cosθ)

But these give opposite sine values, and hence opposite velocities v. (v = –Aω sinθ)

In other words, θ uniquely identifies a particular point in the oscillation cycle.

Note that you can (and will, eventually) have a phase >2π. Add 2π to any phase and you get the same x and v.

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0 1 2 3 4 5 6

Phase (theta), radians

-1

-0.5

0

0.5

1

cos(t

heta

)

θ 2π–θ

0 1 2 3 4 5 6

Phase (theta), radians

-1

-0.5

0

0.5

1

-sin

(theta

)

Phase vs TimeIf we know what the phase was when we started, and we know how quickly the phase is changing, then we can tell what the phase is at any time t.

How quickly is the phase (θ) changing?

• If it goes all the way around and comes back to where it started, it’s changed by 2π.

• It completes one cycle in one period (T).

• Since the ball is going around at a constant speed, we can useΔθ/Δt = 2π/T = 2πf = ω

Where did it start? What value did θ have at t=0? Call that value ϕ.

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P’

θA

x

P

x

ω

light

A–A

Phase ConstantUp until now we’ve always started our oscillator by lifting the mass and letting go -- that is, we’ve started at the “top” of the cycle. That gave us our cosine function.

To start at any point in the cycle, we can use the same function and just change its value at t = 0. We can do this with the phase constant.

x = A cos(ωt + ϕ)

This lets us get any value of x at t = 0:

x0 = A cos(ϕ)43

So, just like for distance d = vt + d0, here we have! θ = ωt + ϕ

Once again, the phase is whatever angle you have to give your trig functions to get the correct x and v (and a).

• Phase is a way to label a point in the cycle.

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Remember that v and a also change throughout the motion; if we’re changing our starting x we have to change our starting v and a as well. The phase constant does this about like you’d expect (you can derive it with calculus):

x = A cos(ωt + ϕ)

v = –Aω sin(ωt + ϕ)

a = –Aω2 cos(ωt + ϕ)

And we can show that this makes sense: remember that for SHM, a = –ω2x. These equations still satisfy this.

Phase Constant in v and a

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θA

Fig. 15-13a

θ = ωt + ϕ

ϕ and A from x0 and v0The physical initial conditions of the system — that is, x0 = x(t = 0) and v0 = v(t = 0) — combined with the properties of the system tell us the phase constant and the amplitude of the oscillations.

Use the general equations of motion we just found, and solve them for ϕ and A. At t=0, we have:

x0 = A cos(ϕ)

v0 = –Aω sin(ϕ)

Two equations, with two unknowns. We can solve this!

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Determining A

Trig identity:

Amplitudein SHM:

Determining ϕ

Phase constantin SHM:

But this isn’t the only correct value of ϕ!! tan(ϕ) = tan(ϕ+π), so ϕ+π is also a possibility.

Plug both ϕ and ϕ+π back into both x0 = A cos(ϕ) and v0 = –Aω sin(ϕ) to determine which to use.

One of two possible answers!

sin vs. cos

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Remember that sin(θ) = cos(θ – π/2)and cos(θ) = sin(θ + π/2).

This means we have our choice of using sin or cos when describing simple harmonic motion. We just have to be consistent about it:

The phase constants (ϕ) differ by π/2.

The signs come from the unit circle, or from calculus: v = dx/dt etc.

x = A cos(ωt + ϕ)v = –Aω sin(ωt + ϕ)a = –Aω2 cos(ωt + ϕ)

orx = A sin(ωt + ϕ)v = Aω cos(ωt + ϕ)a = –Aω2 sin(ωt + ϕ)

So now you know:• what phase is, and what the “phase angle” means.

• what the phase constant, a.k.a. the initial phase, represents, and how to use it.

• how the amplitude A and phase constant ϕ are related to the initial position x0 and initial velocity v0.

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Equations of Motion for SHMWe can now completely describe the motion of a simple harmonic oscillator, starting our clock at any point we like:

x = A cos(ωt + ϕ)v = –Aω sin(ωt + ϕ)a = –Aω2 cos(ωt + ϕ)

We know this motion is the result of a restoring force which looks like F = –kx, giving us a = –(k/m)x.

We know what determines the angular frequency ω, and what determines the amplitude A and phase constant ϕ.

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xmax, vmax, amaxYou can read the maximum displacement (xmax or A), the maximum velocity (vmax), and the maximum acceleration (amax) directly from the equations of motion.

The sine and cosine functions are always between –1 and +1.

x = A cos(ωt + ϕ) v = –Aω sin(ωt + ϕ) a = –Aω2 cos(ωt + ϕ)

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Time

Displacement cos(ωt + ϕ)+1

–1

e.g. Block Cat on a springDiana, Duck of Science!, is studying Simple Harmonic Motion. She puts her assistant Bob, the Thrill-Seeking Cat (mBob!=!4.00!kg), on a platform on a spring (k!=!400!N/m), and sets it in motion.

• What is the largest amplitude of oscillations Diana can use before Bob leaves contact with the platform?

• At what point in the cycle will Bob leave the platform at this amplitude?

• If Bob is oscillating with an amplitude of 5.0!cm, how long will it take him to move from x1!=!5.0!cm to x2 = 2.5!cm?

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Quark!

μ!

Energy in SHM

Consider a cart on a horizontal air track.

• Kinetic energy:

• Spring potential energy:

There’s gravity but it does no work on the system, so it doesn’t contribute to the energy. So this is it for the whole mass-spring system.

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stretch = x only because it’s a horizontal track!

Fig. 15-5

Energy ConservationEnergy can only transfer between the mass (as kinetic energy) and the spring (as potential energy). It can’t go anywhere else. So the total energy is

E = K + U

where E is constant no matter what the mass is doing. We know K and U for a mass on a spring, so we have

E = (1/2)mv2 + (1/2)kx2

Now we need to find total energy E.

How do we do that? Well, E is the same everywhere, so is there anywhere we know the values of x and v?

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Total Energy in SHMDuring simple harmonic motion, v = 0 when x = A. Plug this in:

So our general description of the energy of SHM is:

This tells us what the energies are doing at any point in the motion. It also gives us a relationship between v and x everywhere (without trig :), which can be very useful!

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Energy vs MotionIs what we’ve found out about the energy consistent with the equations of motion?

Since

So our expressions for displacement and velocity are consistent with energy conservation, and we’re not missing something.

Velocity vs Displacement

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Starting with the SHM energy equation:! (1/2)mv2 + (1/2)kx2 = (1/2)kA2

you can rearrange this to find the velocity at a given displacement of the mass:

The ± symbol just means that v can be either positive or negative at any given displacement x — which is consistent with what we already know, that at any position the mass can be going either left or right (or up or down, etc).

Problem Solving StrategiesIdentification: any force that looks like F = –kx causes simple harmonic motion. (Might not be called “k” and “x”!)

If the question involves time, look to the equations of motion. (Remember that the phase constant is for describing things at t = 0, so it’s related to time.)

If the question involves velocity and displacement, look to the energy equation. (Remember that it uses x2 and v2, so you have to figure out the signs from the situation. Be careful!)

If the question involves acceleration and displacement, look at a = –(k/m)x (which comes from F = –kx and F = ma).

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Energy vs time

As time goes on, energy shifts between kinetic and potential, but the total energy is constant.

Fig. 15-6a

Energy vs displacement

Fig. 15-6b

As position changes, energy shifts between kinetic and potential, but the total energy is constant.

A parabolic potential energy curve is another signature of simple harmonic motion.

Energy with Diana!a) What are Diana’s maximum and minimum velocities? What is her maximum accelleration?

b) What are v and a when Diana is halfway to the centre from her original position (i.e. at x = 0.010 m)?

c) What are E, U, and K at this position?

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ice (no friction)

Quark!

k = 200 N/m. m = 0.50 kg.

x0 = 0.020 m. v0 = 0.

So now you know:• how to describe the energy in Simple Harmonic

Motion.

• We used a mass on a spring, but all SHM will look similar! (As you’ll soon see...)

• how the different types of energy change depending on displacement, and on time.

• For time: plug the equations of motion into the energy equations!

• general problem-solving strategy for SHM problems.

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SHM with a Simple PendulumUntil now, we’ve been discussing harmonic oscillators moving in a straight line. We can describe the motion and we know how the energy behaves.

The same principles apply to any harmonic oscillator.

• We may have to use different variables, but we can figure out what these need to be by comparing to the mass & spring system and looking for patterns.

• As long as the (net!) force is proportional to the displacement from some equilibrium position, and points back to equilibrium, you have SHM.

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The Simple PendulumA simple pendulum is a point mass hanging by a massless, unstretchable string.

• This is never technically true in real life, but it’s a good assumption as long as the mass is heavy enough and the string doesn’t stretch much.

We’re hoping we can make this look likeF = –kx, so that we can just take the mass-on-a-spring formulas and replace all the variables. Let’s set it up with that in mind.

• We need a coordinate with units of length. Make x the arc length from the rest position. Then x = Lθ.

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x

L

m

θ

right = positiveleft = negative

right = positiveleft = negative

(radians)66

Pendulum ForcesThe only thing that will affect the motion of the pendulum is the component of the force of gravity which points along the direction of motion: F = –mg sinθ. The string cancels the rest.

This pulls the mass back towards the equlibrium point (θ=0), so it’s a restoring force.

Remember that if θ is in radians, then for small angles sinθ≈θ. Then" F ≈ –mgθ = –mg(x/L)

or mg

θ

T (tension)

mg cosθ

mg sinθ

constant

Minus sign means F points left when θ is on the right, and vice versa.

Pendulum SHM!F = –(mg/L)x looks just like F = –kx, so all of our previous discussion of simple harmonic motion applies.

Just make the replacement k ⇒ (mg/L) and everything will

work. (Provided the pendulum swing is a small angle!)

Example: Pendulum StandardFind the period and frequency of a simple pendulum 1.000 m long, on Earth (g = 9.806 m/s2).

This was used for the original definition of the second!

• Not a good standard — g varies from place to place, and even with the phase of the moon...

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Pendulum EnergyKinetic energy: K = (1/2)mv2, as before.

Potential energy: define h=0 at the bottom of the swing. Then U = mgh. But we want this in terms of arc length x...

69

x

L

m

θ

h

L cosθ

The Binomial TheoremI am very well acquainted, too, with matters mathematical.I can solve equations, both the simple and quadratical.About the binomial theorem I am teeming with a lot of news,With many cheerful facts about the square of the hypotenuse." (“I Am the Very Model of a Modern Major General”," from Gilbert & Sullivan’s “The Pirates of Penzance”.)

You don’t need to memorize (or write down) this, but the Binomial Theorem is often handy:

Consider n = 2, which you should already be familiar with:

Think about what happens if b is very small...

70

(a + b)n = an + nan−1b + . . . + nabn−1 + bn

(a + b)2 = a2 + 2ab + b2

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Let’s work with that square root sign. Since one term (L2) is much larger than the other (x2), we can use:

Here, a = L2, b = –x2, and n = 1/2. Then:

or

So

“much greater than”

So after all that, we found:

Remember that we identified k = (mg/L). Look familiar?

This is the same form as the potential energy for a mass on a spring, with k replaced.

(Remember that this is only true for small swings, i.e. small θ, small x.)

This reinforces what we’ve been saying: all Simple Harmonic Motion setups work exactly the same way, once you define “x” and “k” in Fnet = –kx.

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• how to describe the motion of a simple pendulum.

• how the energy of a simple pendulum behaves.

• how to apply what we learned about SHM from the mass-on-a-spring to any Simple Harmonic Motion situation!

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So now you know:Until now, we’ve been assuming everything is frictionless. In real systems, there’s always some sort of dissipative effect—air resistance, friction, sound generation, etc.

Oscillators left to themselves will decrease in amplitude—their total mechanical energy is being lost to heat due to friction, and E = (1/2)kA2—and they’ll eventually stop. This is called damping.

• The oscillators do not slow down over time; the frequency does not depend on the amplitude!

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Damped Oscillations

undamped

damped

Levels of DampingIf the damping isn’t strong enough to keep the system from bouncing, the system is underdamped.

If the damping is just barely enough to completely prevent oscillations, the system is critically damped.

If the damping is any stronger than that, the system is overdamped.

• A critically damped system will return to equilibrium the fastest, without overshooting. More damping will slow it down.

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Example: Shock absorbers in a car or bike.

• Too much damping and they can’t react to multiple bumps. Too little damping and they bounce (and bottom out more easily). They’re usually made slightly underdamped.

• If your shocks wear out, the fluid in them loses some of its viscosity and you don’t get enough damping. Boing!

Example: many good quality weigh scales have some sort of damping.

• Critical damping is best for a quick, accurate reading.

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Simple DampingLet’s look at the simplest form damping can take: a frictional damping force that’s directly proportional to the speed of the oscillator.

• This is a common form of damping; it’s the form you see when your oscillator is in a viscous fluid.

• Resistive damping in electrical circuits have the same mathematical form, too, it turns out.

So the frictional damping force is Fd = –bv, where b is a constant which describes the degree of damping.

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As before, we can determine the equations of motion (the description of how the oscillator moves in time) from Newton’s Second Law (F = ma).

We need to include the damping force in the total force on the oscillator: Fnet = –kx – bv.

Then Newton’s second law becomes:

This is another differential equation that determines x(t). The solution depends on the degree of damping...

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Finding x(t) with damping

If the system is underdamped (damping is small compared to the spring force), the solution is:

And the angular frequency ω’ is

So if you have two oscillators with the same k and m, but with different damping (different values of b), they will oscillate with different frequencies. But the frequencies will not change over time!

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e = 2.1828... Amplitude is no longer constant!The apostrophe (“prime”) just says it’s not quite the same as the undamped ω.

“envelope”

When ω’=0, there are no oscillations; the cosine is just a constant (cosϕ). So you just return exponentially to equilibrium. How large does b have to be for this to happen? (Call this bc.)

At this point we say the system is critically damped.

For b < bc, the system is underdamped and oscillates we’ve described.

For b > bc, the system is overdamped. Note that ω’ becomes imaginary. We get a different solution:

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(if ω’=0, then (ω’)2=0)

You don’t need to know this formula. The important thing is that x goes exponentially to zero—no oscillating.

bc is not a standard symbol; I made it up. But it’s handy.

x = Ae−b

2m t cos(φ)

b < 2√

km

b = 2√

km

Summary of Damping

b > 2√

km

→ underdamped.

→ critically damped.

→ overdamped.

Angular frequency:

You don’t need to know this formula.

Damping dissipates the mechanical energy (E = K+U) out of the system (to heat!). To see how, let’s look at how the total mechanical energy changes with time.

Recall what F=ma looks like for the damped system:

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Damping and Energy Forced OscillationsIf you take a mass on a spring and start it moving, it will oscillate at some frequency determined by k, m, and b, and it will eventually stop.

But if you attach the other end of the spring to a motor instead of something stationary, you can make the mass oscillate at any frequency you like.

This oscillating push is called a driving force.

Some other examples:

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• Vibrating motor

• Wave pool

• Speaker

• Bored student in a swivel chair

The amplitude with which the mass oscillates will depend on the driving frequency.

You’d expect the oscillator to react more strongly at its “natural frequency”—the frequency it would oscillate at if you just pulled it and let go.

• Driving at this frequency means you’re pushing down when the oscillator is moving down, and pulling up when the oscillator is moving up. When you reinforce the motion like this you get bigger motion. This is resonance.

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Resonance

Video from University of Salford (UK).http://www.acoustics.salford.ac.uk/feschools/waves/wine3video.htm

Loudspeaker, generating a very loud sound at the glass’s resonant frequency.

(Very, very slow motion.)

Examples of Resonance

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Weird car rattles and other noises, that only occur at certain engine speeds.

Parts of the bus rattling or buzzing, again at specific speeds.

Tuning in a radio—only a specific frequency (well, a narrow range) gets picked up.

Earthquake-proofing a building—changing its resonance frequency to be nowhere near earthquake frequencies.

The Millenium Bridge in London:http://www.youtube.com/watch?v=gQK21572oSU

The Tacoma Narrows Bridge:http://www.youtube.com/watch?v=xox9BVSu7Ok

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Sinusoidal Driving ForceSince we’ve been looking at sinusoidal oscillation, let’s consider a sinusoidally-varying force, such as:

! F(t) = Fmax cos(ωd t)

• The force doesn’t have to look like this, of course, as long as it oscillates. But this is an interesting case to study because it’s got the same form as simple harmonic motion—and it’s a fairly common form of driving force, too.

Let’s also assume that we’re working with a simple harmonic oscillator, because it’s interesting.

We can set up the F = ma equation again, as before, including this force as well as the restoring and damping forces:

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Hardest the driving force pushes

Angular frequency of the driving force

Damping force

Spring force

Driving force

Amplitude vs FrequencyBy substituting for x and v, and some math, can find out the amplitude of the forced oscillations:

A is maximum near where k – mωd2 = 0, or

At low frequency (ωd→0), A = Fmax/k (Hooke’s Law!).

At high frequency (ωd→∞), A = 0.

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(look familiar?)

You don’t need to know this formula.

Resonance frequency! is the frequency at which the mass on a spring would oscillate if you were to nudge it and let it go. For this reason, this frequency is often called the natural frequency of the system.

Since if you were to drive the system at this frequency it would resonate (oscillate at very large amplitude), it’s also often called the resonance frequency (or “resonant frequency”, depending on who you ask).

How would you find the resonance frequency of a pendulum?

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�k/m

Damping and Resonance

So now you know:• what damping means to an oscillator.

• how different amounts of damping affect an oscillator.

• what “simple” damping is.

• how damping affects the mechanical energy of an oscillator over time.

• how the amplitude of forced oscillations depends on the frequency of the driving force.

• what resonance is, and how it depends on the properties of the system (k, m, b, etc.).

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