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Physics Part IA Cambridge University, 2012-2013

Notes for Oscillating Systems (Part 1) Undamped simple harmonic motion: This is an ideal case of simple harmonic motion in which the only force acting on an object is a restoring force towards equilibrium: By Newton's Second Law:

x m

[?] x +

Equilibrium position

Letting

-kx=m x

k x =0 m

Which is the equation of motion for an

2 k /m=o0 such that the equation of motion is not specific to the mass/spring

system: 2

[?] x +o 0 x=0 Which is the equation of motion for undamped SHM in general. The solutions to this differential equation can be expressed in two forms: Complex exponential form: Let

at

2 at

x= A e [?] x = A a e

Real (physical) sinusoidal form:

:

[?] A e at ( a 2 +o 20) =0, A e at [?] 0

[?] a =+- io 0

[?] x= A e

R A +B Im Re

ph

a0

a0

i o0 t

-i o 0 t

+B e

Physics Part IA Cambridge University, 2012-2013

ph

A-B ()i

R

Due to the equivalence of exponentials and sinusoids:

x=( A+ B ) cos o 0 t + ( A-B ) isin o 0 t : A

Where

B

and

are in general

complex.

o0

We are only interested in the real part of this, so we are looking for the real parts of

A + B and ( A-B ) i :

defines the period of the motion and

is thus called the 'angular frequency':

a0 =

If

amplitude and

ph=

phase,

then:

2p

2p

T = - o 0=

o0 T

When

t=0 :

x= A+ B

[?] o 0=2 pu

[?] R { A+ B } =a0 cos ph

Where

u is the oscillatory frequency of

the motion and

T

When

is the oscillatory

t=p /2 o0

period.

:

x=( A-B ) i

[?] R {( A+ B ) i }=a0 sin ph

[?] R { x }=a0 cos ph cos o0 t+ a0 sin ph sin o0 t

[?] R { x }=a0 cos(o 0 t + ph) Or

R { x }=C cos o0 t+ D sin o 0 t

a0 =[?] C 2 + D2 , ph=tan -1

-D C

Mechanical systems displaying undamped SHM: All mechanical systems displaying SHM have the defining equation

o20

is defined differently for each:

x + o20 x=0 , but

Physics Part IA Cambridge University, 2012-2013 Simple pendulum:

-F T =m s =ml th

th

s

s l mg FT

[?]-mg sinth=ml th

As long as

[?]-gth=l th

g x=0

[?] th+

l

[?] o 20=

g l

[?] o 20=

t

I th

FT G Torsional pendulum:

G=I th=-tth

t x=0

[?] th+

I th is small, sin th [?] th :

Physics Part IA Cambridge University, 2012-2013

Velocity, acceleration and phasor diagrams: Phasor diagrams represent the phase relationships between displacement, velocity and acceleration for an oscillating system: Displacement:

Velocity:

x=a 0 cos (o 0 t+ ph)

x =

Phase relationship:

Acceleration:

d ( x ) =-a0 o0 sin(o 0 t + ph) dt

Phase relationship:

x=a 0 cos (o 0 t+ ph)

(

x =a 0 cos o 0 t + ph+

2 x = d 2 ( x )=-a0 o20 cos (o 0 t+ph) dt

Phase relationship:

p

2 )

x =a 0 cos ( o0 t + ph+ p )

The successive phase difference between displacement, velocity and acceleration is therefore

p /2 :

Argand plane:

a0 e

Phasor diagram:

i (o0 t +ph )

a0

Im

a0

o0 t a0 cos (o0 t+ph)

a0

a0

ph

o0 t

Re

a0 cos (o0 t+ph)

a0

ph

x Physics Part IA Cambridge University, 2012-2013 Velocity: Argand plane:

Phasor diagram:

Im Re

a0 o 0

a0

(

a0 o 0 cos o 0 t + ph+

a0 o 0 e

(

i o0 t +ph +

p

2 p

2 )

p

2 )

)

a0

x a0 o 0

a0

(

a0 o 0 cos o 0 t + ph+

Physics Part IA Cambridge University, 2012-2013 Acceleration: Argand plane:

Phasor diagram:

a0 o 20 cos ( o0 t + ph )

x Re Im 2

a0 o 0

a0 a0 o 20 cos ( o0 t + ph ) a0 o 20 e

i (o 0 t +ph )

a0

Superposition of simple harmonic motion: The equation of motion for SHM is linear, hence solutions of this equation can be superimposed to give another solution: This can be proven for the general case of superimposing two simple harmonic oscillations of the same frequency;

SHM 1

and

SHM 2

:

Physics Part IA Cambridge University, 2012-2013

SHM 1 : x1+ o0 x 1=0, SHM 2 : x2+ o0 x 2=0 Where

x 1+ x 2=x [?] x1+ x2= x

:

l1 SHM 1+ l2 SHM 2= l1 ( x1 +o 0 x 1 ) + l 2 ( x2 +o0 x 2 )

( l1 + l 2 ) x + ( l 1+ l2 ) o0 x

l3 x + l3 o0 x The main use of phasor diagrams is to superimpose SHMs of either common or different frequency.

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