Simple Harmonic Motion

16th October 2008

Quick review…

(Nearly) everything we’ve done so far in this course has been about a single equation

F ma=a = constant

210 2

( )x t v t at= +a = f(t)

( ) ( )d dx t f t t t= ò

a is central

Circular motion…

What we’re doing now…

Consider new type of force, to extend the types of situations we can deal with

F kx= -

ma kx

ß

= -

Example… Particle on a spring

ˆspring

kx= -F ir

2

2

d

d

xma m kx

t= = -

Solutions…

You saw yesterday that one solution to this problem was

( ) cosk

x t A tm

æ ö÷ç ÷ç= ÷ç ÷ç ÷çè ø

However, this is not the only solution… It turns out that the following solution also

works

( ) sink

x t B tm

æ ö÷ç ÷ç= ÷ç ÷ç ÷çè ø

Most general solution

It turns out that the most general solution is given by (see 18.03…)

( ) cos sink k

x t A t B tm m

æ ö æ ö÷ ÷ç ç÷ ÷ç ç= +÷ ÷ç ç÷ ÷ç ç÷ ÷ç çè ø è ø

km

w=2

Tpw

= Ü

Angular frequency

Period

What is the period?

Consider the oscillator at t = 0

( ) ( )( ) cos sinx t tt A Bw w= +

( ) ( )(0) cos 0 sin 0x A B= +

(0)x A=

Consider the oscillator at t = T = 2/2 2

( ) cos sinxT A Bp p

w ww w

æ ö æ ö÷ ÷ç ç÷ ÷= +ç ç÷ ÷ç ç÷ ÷ç çè ø è ø

( ) ( )( ) cos 2 sin 2xT A B Ap p= + =

What does the motion look like?

( ) ( )( ) cos sinx t A t B tw w= +

T

Velocity and acceleration

We can differentiate our expression to find velocity and acceleration

( ) ( )

( ) ( )

( ) ( )2 2 2

( ) cos sin

( ) sin cos

( ) cos sin ( )

x t A t B t

v t A t B t

a t A t B t x t

w w

w w w w

w w w w w

= +

= - +

= - - = -

How to find A and B?

Use initial conditions

Since there are two constants, we need two initial conditions

These can be anything, but let’s try an example with initial position x(0) and

initial velocity v(0)

( ) ( )( ) cos sinx t A t B tw w= +

Initial Position

( ) ( )( ) cos sinx t tt A Bw w= +

If t = 0

( ) ( )( ) ( )

(0) cos sin

(0) cos

0

sin

(0

0 0

)

0x A B

x A B

x A

w w

+

´ +

=

´=

=

So the constant A is simply equal to the original displacement

Initial velocity

( ) ( )( ) cos sinx t A t B tw w= +

To find the velocity, we need to differentiate

( ) ( )d ( )( ) sin cos

dx t

v A Bt

tt tw ww w= = - +

At t = 0

(0)v Bw=

And so B is equal to (0) /v w

Summary

We’ve learnt how to deal with forces like

kF kx a x

m= - Þ = -

We found that the solution looks like

( ) ( )( ) cos sin /x t A t B t k mw w w= + =

We can use initial conditions to find A and B

We can use this expression to find out anything we want about the motion