Simple Harmonic Motion 16 th October 2008. Quick review… (Nearly) everything we’ve done so far...
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Transcript of Simple Harmonic Motion 16 th October 2008. Quick review… (Nearly) everything we’ve done so far...
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