@kul 2 shm - circular - energi

8/12/2019 @Kul 2 SHM - Circular - Energi

1/31

SHM and Circular Motion

Uniform circularmotion projected onto

one dimension issimple harmonic

motion.


2/31

SHM and Circular Motion

x(t) Acos

d

dt

t

x(t) Acos t

Start with the x-component ofposition of the particle in UCM

End with the same result asthe spring in SHM!

Notice it started at angle zero


3/31

Initial conditions:

t 0

We will not always start ourclocks at one amplitude.

x(t) Acos t 0 v x (t) Asin t 0 v x (t) v max sin t 0


4/31


5/31

Phase Shifts:


6/31

An object on a spring oscillates with a period of 0.8s andan amplitude of 10cm. At t=0s, it is 5cm to the left of

equilibrium and moving to the left. What are its positionand direction of motion at t=2s?

x(t) Acos t 0

x 0 5cm Acos 0 Initial conditions:

0 cos 1 x 0

A

cos

1 5cm10cm

120

23

rads

From the period we get: 2T

20.8s

7.85rad/s


7/31

An object on a spring oscillates with a period of 0.8s andan amplitude of 10cm. At t=0s, it is 5cm to the left of

equilibrium and moving to the left. What are its positionand direction of motion at t=2s?

x(t) Acos t 0

7.85rad /s0 2

3 rads

A 0.1m

t 2s

x(t) 0.1cos 7.85 2 23

x(t) 0.05m


8/31

We have modeled SHM mathematically.Now comes the physics.

Total mechanical energy is conservedfor our SHM example of a spring with

constant k, mass m, and on africtionless surface.

E K U 12

mv 2 12

kx 2

The particle has all potential energyat x=A and x= A, and the particlehas purely kinetic energy at x=0.


9/31

At turning points:

E U 12

kA 2

At x=0:

E k 12

mv max2

From conservation:12

kA 2 12

mv max2

Maximum speed as related toamplitude:

v max k

mA


10/31

From energy considerations:

From kinematics:

Combine these:

v max k

mA

v max A

k

m

f 12

k m

T 2 m

k


11/31

a 500g block on a spring is pulled a distance of 20cm andreleased. The subsequent oscillations are measured to

have a period of 0.8s. at what position or positions is theblocks speed 1.0m/s?

The motion is SHM and energy is conserved.

1

2mv 2

1

2kx 2

1

2kA 2

kx 2 kA 2 mv 2

x A 2 mk

v 2

x A2 v

2

2

2

T

2

0.8s

7.85rad/s

x 0.15m


12/31Dynamics of SHM

Acceleration is at a maximum when the particle is atmaximum and minimum displacement from x=0.

a x dv x (t)dt d Asin t dt

2Acos t



Acceleration isproportional to the

negative of thedisplacement.

a x 2Acos t

a x 2x

x Acos t



As we found with energyconsiderations:

a x 2x

F ma x kxma x kx

a x k mx

According to Newtons 2 nd Law: a x d

2xdt 2

Acceleration is notconstant:

d 2xdt 2

k m

x

This is the equation ofmotion for a mass on aspring. It is of a general

form called a second orderdifferential equation.


15/31

2 nd -Order Differential Equations:

Unlike algebraic equations, their solutions are notnumbers, but functions.

In SHM we are only interested in one form so we canuse our solution for many objects undergoing SHM.

Solutions to these diff. eqns. are unique (there is onlyone). One common method of solving is guessing the

solution that the equation should have

d 2x

dt2

k

mx

Fromevidence, we

expect thesolution:

x Acos t 0


16/312nd

-Order Differential Equations:

Lets put this possible solution into our equation andsee if we guessed right!

d2

xdt 2

k m

x

IT WORKS. Sinusoidal oscillation ofSHM is a result of Newtons laws!

x Acos t 0

d 2xdt 2

2Acos t

dx

dt

Asin t

2Acos t k m Acos t 2 k

m


17/31

What about vertical oscillations ofa spring-mass system??

Fnet k L mg 0Hanging at rest:

k L mg

L mk

g

this is the equilibriumposition of the system.


18/31

Now we let the systemoscillate. At maximum:

But:

Fnet k L y mg

Fnet k L mg ky

k L mg 0So:

Fnet ky

Everything that we have learned abouthorizontal oscillations is equally valid for

vertical oscillations!


19/31

The Pendulum

Fnet t mg sin ma td 2sdt 2 gsin

Equation of motionfor a pendulum

s L


20/31

Small Angle Approximation:

d 2sdt 2

gsin

When is about0.1rad or less, h

and s are about thesame.

sin cos 1tan sin 1

d 2sdt 2

g

sL

Fnet t md 2sdt 2

mgsL

f


21/31

The Pendulum

Equation ofmotion for a

pendulum

d2

sdt 2

gsL

gL

(t) max cos t 0 x(t) Acos t 0

h l h d l ll h d f l


22/31

A Pendulum Clock

What length pendulum will have a period of exactly 1s?

g

L

T 2 L

g

g T

2

2

L

L 9.8m/s2 1s

2

2

0.248m


23/31

Conditions for SHM

Notice that all objects thatwe look at are described

the same mathematically.

Any system with a linear restoring

force will undergo simpleharmonic motion around theequilibrium position.


24/31

A Physical Pendulum

d 2

dt 2 mgl

I

I mgd mglsin

when there ismass in the

entire pendulum,not just the bob.

Small Angle Approx.

mglI


25/31

Damped Oscillations

All real oscillators are dampedoscillators. these are any that slow

down and eventually stop.a model of drag force for

slow objects:

Fdrag bv

b is the damping

constant (sort of like acoefficient of friction).


26/31

Damped Oscillations

F Fs Fdrag kx bv ma

kx bdxdt

md 2xdt 2

0

Another 2 nd -order diff eq.

Solution to 2 nd -order diff eq:

x(t) Ae bt / 2m

cos t 0

k m

b 2

4m 2

02 b2

4m 2


27/31

Damped Oscillations

x(t) Ae bt / 2m cos t 0

A slowly changing linethat provides a border to

a rapid oscillation is

called the envelope ofthe oscillations.


28/31
http://www.youtube.com/watch?v=IqK2r5bPFTM&feature=related


29/31

DrivenOscillations

Not all oscillating objects are disturbed from restthen allowed to move undisturbed.

Some objects may be subjected to a periodicexternal force.


30/31

DrivenOscillations

All objects have a natural frequency at whichthey tend to vibrate when disturbed.

Objects may be exposed to a periodic force witha particular driving frequency .

If the driven

frequency matchesthe naturalfrequency of an

object, RESONANCE occurs


31/31

THE

END

@kul 2 shm - circular - energi

Documents