Simple Harmonic Motion2

8/3/2019 Simple Harmonic Motion2

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400grams

200grams

F

O

R

C

E

(N)

EXTENSION (e)

Gradient = spring constant

600grams

\

Extension of spring


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O

To

mg

mg

To = 0

mg = To

mg = ke

ge = kme

T1

mg

x mg T1 = ma

mg

k(e+x) = mamg ke kx = ma

0kx = ma

a = k x

m


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a = k xm

a = 2xcompare

2 =km

=km

= ge

= 2f

= 2T

T = 2

T = 2 m

k

T = 2 eg

Independentof a :

On the moon,

T is stillthe same


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Small masses vibrate with

shorter periods

Large masses vibrate with

longer periods


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Springs with smaller constants

vibrate with longer periods

Springs with larger constants

vibrate with shorter periods


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O

O

EqulibriumSmooth surfaceNo Tension

A

Displaced to AOscillatingAmplitude= OA = OB

B


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F = ma

-T1 = ma

O AB

x

aT1

-kx = ma

a = - xkm

a = 2xcompare

2 =km

=km

= 2f

= 2T

T = 2

T = 2 m

k


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A pendulum consists of asmall bob of mass m,suspended by a light

inextensible thread oflength l, from a fixedpoint.

The bob can be made tooscillate about point O ina vertical plane along thearc of a circle.

We can ignore the mass

of the thread


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In order to be in SHM, the restoring force mustbe proportional to the negative of thedisplacement. Here we have:

which is proportional to sin and not to itself.


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DisplacementRestoring

force


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However, if the angle is small,sin .


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ma = - mg sin

a = - g

L

x

sin = xL

= x

L

mg

a = - g xL

a=2xcompare

2

=

g

L

= gL


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ma = - mg sin

a = - g

a = - g xL

a=2xcompare

2

=

g

L

= gL = 2f

= 2T

T = 2

T = 2 Lg


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SHORT PENDULUMS

HAVE A SHORT

PERIOD OF

OSCILLATION

LONG PENDULUMS

HAVE A LONG

PERIOD OF

OSCILLATION


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GRAVITY ON EARTH9.8 M/S2 GRAVITY ON MOON1.6 M/S2

STRONGER GRAVITY FIELDS

RESULT IN SHORTER PERIODS

OF OSCILLATION


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Solving SHM ProblemsA pendulum 1.00 meters long oscillates at 30 times a minute.

What is the value of gravity ?

30 oscillations per minute = 30 / 60 = 0.50 oscillationsper second

f = 1/ T, f = 1 / 0.50 = 2.0 sec T = 2 (l / g)1/2 , g = 4 2 l / T2 g = 4 2 x 1.00 / 2.02 = 9.87 m/s2

T = 2 l / g


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compare

S l in SHM P bl


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Solving SHM ProblemsA torsion pendulum consists of a 2.0 Kg solid disk of 15 cm

radius. When a 1 N-m torque is applied it displaces 150. What

is its frequency of vibration?

The moment of inertia of a disk is given by: I = mr2

I = (2.0) (.15)2 = 0.0225 Kg-m2 (note: 15 cm = .15m)

= - k , k = / , (150 in radians = 0.262 rad)

k = 1 / 0.262 = 3.82 N-m / radian

T = 2 (0.0225 / 0.262)1/2 = 0.482 seconds f = 1 / T = 1 / 0.482 = 2.07 hertz

= - k


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T = 2 Lg

T2

= Lg

T2

42=

Lg

g = 42 LT2

T2 = L42

g

constant

T2 L

T2

L0

Gradient = T2

L

= 42

g

g = 42 .

Gradient


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Experiment

Using a long clamp stand, apendulum bob, some light string anda stop watch to investigate therelation ship between g, l and T

For a pendulum of known lengthcount the time taken for 10 completeoscillations (there and back).

Use the pendulum formula tocalculate the force of gravity.

Repeat with 3 other lengths.


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Initially :At t = 0, particle at x

x

After time t : = t , particle at P

PQ

P is projected to Q:

O

y

r

sin = yr

y = r sin t


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As P movesround acompletecircle,Q moves fromO to A, then

from A to Band finallyback to O.

A

B

O


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Amplitude ofOscillation of Q= radius

A

B

OPeriod ofOscillation of Q

= Period of theCircular Motionof P


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y = r sin t

v = dydt

= r cos t

a = d2ydt2

= - 2r sin t

= - 2y


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x

t0

xo

-xo


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Damped Oscillations

Damped oscillation is harmonic motion with africtional or drag force. If the damping issmall, we can treat it as an envelope that

modifies the undamped oscillation.


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However, if the damping is

large, it no longer resemblesSHM at all.

A: underdamping: there area few small oscillationsbefore the oscillator comesto rest.

B: critical damping: this is the fastest way to get to

equilibrium.C: overdamping: the system is slowed so muchthat it takes a long time to get to equilibrium.


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Forced Vibrations; Resonance

Forced vibrations occur when there is a

periodic driving force. This force may or maynot have the same period as the naturalfrequency of the system.

If the frequency is the same as the naturalfrequency, the amplitude becomes quite large.This is called resonance.


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Examples of resonance.

Pushing a child on a swing maximum Awhen pushing = oTuning a radio electrical resonance occurswhen o of tuning circuit adjusted to match of incoming signalPipe instruments - column of air forced tovibrate. If reed = o of column loud soundproduced

Rotating machinery e.g. washing machine.An out of balance drum will result in violentvibrations at certain speeds


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Bartons Pendulum

All objects have a natural frequency ofvibration or resonant frequency. If you forcea system - in this case a set of pendulums -to oscillate, you get a maximum transfer ofenergy, i.e. maximum amplitude imparted.

When the driving frequency equals theresonant frequency of the driven system.The phase relationship between the driver

and driven oscillator is also related by theirrelative frequencies of oscillation.


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You also get a very clear illustration of thephase of oscillation relative to the driver.

The pendulum at resonance is /2 behindthe driver, all the shorter pendulums arein phase with the driver and all the longerones are out of phase.

The amplitude of the forced oscillationsdepend on the forcing frequency of the

driver and reach a maximum when forcingfrequency = natural frequency of thedriven cones.



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The sharpness of the

resonant peak dependson the damping. If thedamping is small (A), itcan be quite sharp; if

the damping is larger(B), it is less sharp.

Like damping, resonance can be wanted orunwanted. Musical instruments and TV/radioreceivers depend on it.


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The amplitude depends on the degree of damping

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Simple Harmonic Motion2