a2-h-41c-shm&oscillations

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4.1c Further Mechanics

SHM & OscillationsBreithaupt pages 34 to 49

January 4 th 2012

CLASS NOTES HANDOUT VERSION


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AQA A2 SpecificationLessons Topics

1 to 3 Simple harmonic motion Characteristic features of simple harmonic motion.Condition for shm: a = (2 f)2 x

X = A cos 2 f tand v = 2f (A2 x 2)Graphical representations linking x, v, a and t .Velocity as gradient of displacement-time graph.Maximum speed = 2fA. Maximum acceleration = (2 f )2 A.

4 to 6 Simple harmonic systems Study of mass-spring system. T = 2(m / k) Study of simple pendulum. T = 2(l / g) Variation of Ek, Ep and total energy with displacement, and with time.

7 to 9 Forced vibrations and resonance Qualitative treatment of free and forced vibrations.Resonance and the effects of damping on the sharpness of resonance.Phase difference between driver and driven displacements.Examples of these effects in mechanical systems and stationary wave

situations


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Oscillations (definitions) An oscillation is a repeatedmotion about a fixed point.

The fixed point, known asthe equilibrium position , is

where the oscillating objectreturns to once theoscillation stops.

The time period, T of anoscillation is the time takenfor an object to perform onecomplete oscillation. equilibrium position


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The amplitude, A of anoscillation is equal to themaximum value of thedisplacement, x

Frequency, f in hertz is equal tothe number of completeoscillations per second.

also: f = 1 / T

Angular frequency, inradians per second is given by:

= 2 for = 2 / Tequilibrium position


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Simple Harmonic Motion

Many oscillating systems undergo a pattern of oscillationthat is, or approximately the same as, that known asSimple Harmonic Motion (SHM).

Examples(including some that are approximate):

A mass hanging from the end of a springMolecular oscillations

A pendulum or swing

A ruler oscillating on the end of a benchThe oscillation of a guitar or violin stringTidesBreathing


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The pattern of SHM motion The pattern of SHM is the same as the side view of anobject moving at a constant speed around a circular path

The amplitude of theoscillation is equal tothe radius of the circle.

The object moves quickest asit passes through the centralequilibrium position.

The time period of the

oscillation is equal to thetime taken for the object tocomplete the circular path.

+ A

- A


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Conditions required for SHM

When an object is performing SHM:

1. Its acceleration is proportional to its displacementfrom the equilibrium position.

2. Its acceleration is directed towards theequilibrium position.

Mathematically the above can be written: a = - k xwhere k is a constant and the minus sign indicates that theacceleration, a and displacement, x are in oppositedirections


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Acceleration variation of SHM

The constant k is equal to(2 f ) 2 or 2 or (2 /T) 2

Therefore:

a = - (2 f ) 2 x (giv en on d ata sh eet)

ora = - 2 x

ora = - (2 /T) 2 x

x

a

+ A- A

- 2 x

+ 2 x

gradient = - 2 or - (2f )2


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Question A body oscillating with SHM has a period of 1.5s and

amplitude of 5cm. Calculate its frequency and maximumacceleration.


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Displacement equations of SHM

The displacement, x of the oscillating objectvaries with time, t according to the equations:

x = A cos (2 f t ) given o n d ata sh eet or

x = A co s ( t )or

x = A co s ( ( 2 / T ) t )

Note: At time, t = 0, x = +A


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Question A body oscillating with SHM has a frequency of 50Hz and

amplitude of 4.0mm. Calculate its displacement andacceleration 2.0ms after it reaches its maximumdisplacement.


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Velocity equations of SHM

The velocity, v of an object oscillating with SHMvaries with displacement, x according to theequations:

v = 2f (A2 x 2) giv en on d ata sh eetor

v = (A2 x 2)or

v = ( 2 / T) (A2 x 2)


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Question A body oscillating with SHM has a period of 4.0ms and amplitude of30 m. Calculate (a) its maximum speed and (b) its speed when its

displacement is 15 m.


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Variation of x, v and a with time

The acceleration, ax depends on the resultant force, Fspring on the mass. Note that the acceleration ax is always in the opposite direction to thedisplacement, X .


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SHM time graphs

T/4 T/2 3T/4 T 5T/4 3T/2

x = A cos (2 f t)

a = - (2 f )2 A cos (2 f t) v = - 2 f A sin (2 f t)

+ A

- A

x

va

time

v m ax = 2 f A a m ax = (2 f )2 A

Note: The velocity curve is the gradient of the displacement curveand the acceleration curve is the gradient of the velocity curve.

a

v

x


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SHM summary table

Displacement Velocity Acceleration

+ A 0 - (2 f )2 A

0 2 f A 0

- A 0 + (2 f )2 A


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SHM graph question

T/4 T/2 3T/4 T 5T/4 3T/2

a

time

The graph below shows how the acceleration, a of an objectundergoing SHM varies with time. Using the same time axis show

how the displacement, x and velocity, v vary in time.


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The spring-mass system If a mass, m is hung from a

spring of spring constant, k and set into oscillation thetime period, T of theoscillations is given by:

T = 2(m / k)

AS Remind er: Spring Constant , k :This is the force in newtons required to cause a changeof length of one metre.

k = F / Lunit of k = Nm -1

Mass on spring - Fendt
http://www.walter-fendt.de/ph14e/springpendulum.htmhttp://www.walter-fendt.de/ph14e/springpendulum.htm


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The simple pendulum A simple pendulum consists of:

a point mass undergoing small oscillations (less than 10 ) suspended from a fixed support by a massless, inextendable thread of length, L within a gravitational field of strength, g

The time period, T is given by:

T = 2(L / g)

Simple Pendulum - Fendt
http://www.walter-fendt.de/ph14e/pendulum.htmhttp://www.walter-fendt.de/ph14e/pendulum.htm


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Question Calculate: (a) the period of a pendulum of length 20cm on

the Earths surface (g = 9.81ms-2 ) and (b) the pendulumlength required to give a period of 1.00s on the surface of

the Moon where g = 1.67ms -2 .


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Free oscillation A freely oscillating objectoscillates with a constantamplitude.

The total of the potential and

kinetic energy of the objectwill remain constant.EP + E T = a constant

This occurs when there are nofrictional forces acting on theobject such as air resistance.


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Question A simple pendulum consists of a mass of 50g attached to the end of athread of length 60cm. Calculate: (a) the period of the pendulum(g = 9.81ms -2 ) and (b) the maximum height reached by the mass if themasss maximum speed is 1.2 ms -1.


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Energy variation with an oscillating spring

The strain potentialenergy is given by:

EP = k x 2

Therefore the maximumpotential energy of anoscillating springsystem= k A 2

= Total energy of thesystem, ET

But: ET = E P + E K k A 2 = k x 2 + E K

And so the kineticenergy is given by:

EK = k A 2 - k v 2

EK = k (A 2 - x 2)


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Energy versus displacement graphs The potential energy curve is

parabolic, given by:EP = k x 2

The kinetic energy curve is aninverted parabola, given by:

EK = k (A 2 - x 2)

The total energy curve is ahorizontal line such that:

ET = EP + EK

- A 0 + A

energy

displacement, x

EP

EK

ET


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Energy versus time graphs Displacement varies with timeaccording to:

x = A cos (2 f t) Therefore the potential energycurve is cosine squared,given by:EP = k A 2 cos 2 (2 f t)

k A 2 = total energy, E T.and so: EP = E T cos 2 (2 f t) Kinetic energy is given by:EK = ET - EPEK = ET - ET cos 2 (2 f t) EK = ET (1 - cos 2 (2 f t) )EK = E T sin 2 (2 f t) EK = k A 2 sin 2 (2 f t)

ET

0 T/4 T/2 3T/4 T

energy

time, t

kA 2

EP

EK


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Damping Damping occurs when frictionalforces cause the amplitude of anoscillation to decrease.

The amplitude falls to zero with theoscillating object finishing in its

equilibrium position.The total of the potential and kineticenergy also decreases.

The energy of the object is said tobe dissipated as it is converted tothermal energy in the object and itssurroundings.


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Types of damping 1. Light DampingIn this case the amplitude graduallydecreases with time.

The period of each oscillation will

remain the same.

The amplitude, A at time, t will begiven by: A = A 0 exp (- C t)where A0 = the initial amplitude andC = a constant depending on thesystem (eg air resistance)


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2. Critical DampingIn this case the systemreturns to equilibrium,

without overshooting, in theshortest possible time after ithas been displaced fromequilibrium.

3. Heavy DampingIn this case the systemreturns to equilibrium moreslowly than the criticaldamping case.

light damping

heavy damping

critical damping

displacement

time

A 0


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Forced oscillations All undamped systems of bodies have a frequency with which

they oscillate if they are displaced from their equilibrium position.This frequency is called the natural frequency, f 0 .

Forced oscillation occurs when a system is made to oscillate by a periodic force. The system will oscillate with the appliedfrequency, f A of the periodic force.

The amplitude of the driven system will depend on:1. The damping of the system.

2. The difference between the applied and natural frequencies.


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Resonance

The maximum amplitude occurs when theapplied frequency, f A is equal to the naturalfrequency, f 0 of the driven system.

This is called resonance and the naturalfrequency is sometimes called the resonantfrequency of the system.


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Resonance curves

applied force frequency, f A

amplitude of drivensystem, A

drivingforce

amplitude

f 0

more damping

light damping

very light damping


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Notes on the resonance curves If damping is increased then the amplitude of the drivensystem is decreased at all driving frequencies.

If damping is decreased then the sharpness of the peakamplitude part of the curve increases.

The amplitude of the driven system tends to be:- Equal to that of the driving system at very low

frequencies.

- Zero at very high frequencies.- Infinity (or the maximum possible) when f A is

equal to f 0 as damping is reduced to zero.


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Phase difference The driven systems oscillations are always behind those

of the driving system.

The phase difference lag of the driven system dependson:

1. The damping of the system.2. The difference between the applied and naturalfrequencies.

At the resonant frequency thephase difference is /2 (90 )


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Phase difference curves

applied force frequency, f A

phase difference of driven systemcompared with driving system f 0

more damping

less damping

- 90

- 180


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Examples of resonance Pushing a swing

Musical instruments (egstationary waves onstrings)

Tuned circuits in radiosand TVs

Orbital resonances ofmoons (eg Io and Europaaround Jupiter)

Wind driving overhead

wires or bridges (TacomaNarrows)


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The Tacoma Narrows Bridge Collapse

YouTube Videos:

http://www.youtube.com/watch?v=3mclp9QmCGs 4 minutes with commentaryhttp://www.youtube.com/watch?v=IqK2r5bPFTM&feature=related

3 minutes newsreel footagehttp://www.youtube.com/watch?v=j-zczJXSxnw&feature=related

6 minutes - music background only

An example of resonance

caused by wind flow.

Washington State USA,November 7th 1940.
http://www.youtube.com/watch?v=3mclp9QmCGshttp://www.youtube.com/watch?v=IqK2r5bPFTM&feature=relatedhttp://www.youtube.com/watch?v=j-zczJXSxnw&feature=relatedhttp://www.youtube.com/watch?v=j-zczJXSxnw&feature=relatedhttp://www.youtube.com/watch?v=j-zczJXSxnw&feature=relatedhttp://www.youtube.com/watch?v=j-zczJXSxnw&feature=relatedhttp://www.youtube.com/watch?v=IqK2r5bPFTM&feature=relatedhttp://www.youtube.com/watch?v=3mclp9QmCGs

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