SHM

Could someone help show me how to derive the equations and when they are used:

ax=-(2pif)^2x
x=Acos(2pift)

thanks.

It might help if you use Latex for equations so that they are more readable.
http://www.thestudentroom.co.uk/wiki/LaTeX

Alternatively, use your Windows character map to find the Greek letters
αβγδεζηθλμπρφω for example
and other useful characters such as squared ² and half ½ .

Or use the superscript and subscript buttons in the word processing window you type your message in. On mine they are directly under the bold and italic buttons. If they are not there you will find the option in your TSR account settings.

Is the first equation meant to be
$a_x=-(2\pi f)^2 x$
?
(done using Latex)

or
a_x = -(2πf)²x
(using the subscript superscript buttons.)

To see how it's done reply to my post using the quote button and take a look at my post.

In answer to your question, the 1st formula isn't derived it's the definition of SHM except that the constant in the defining equation is ω² and ω is by definition 2πf

The 2nd equation is obtained by integrating the first one twice.
However, at A level in physics it's normally shown to be true by differentiating it twice to obtain the 1st equation.

Oh, why is it Cos(ω)?

Could someone help show me how to derive the equations and when they are used:

ax=-(2pif)^2x
x=Acos(2pift)

thanks.

It's A cos(ωt) in the example.
This formula for displacement can be either
x=Asin(ωt) or x=Acos(ωt) depending on where you consider time starts.
If it starts when the object is at maximum displacement from equilibrium you can use cos. If you consider it starts from the equilibrium position, where x=0, then you can use sin. It just depends where the object is when t=0

Just in case you wanted to know how the expression for the displacement of an object moving in simple harmonic motion is derived:

$a = -\omega^2 x$

Acceleration is a function of displacement, so $a = v \dfrac{dv}{dx}$

$v \dfrac{dv}{dx} = -\omega^2 x$

$\displaystyle \int v \ dv = -\omega^2 \int x \ dx$

$\dfrac{1}{2} v^2 = - \dfrac{1}{2} \omega^2 x^2 + C$

We know that when the velocity of the object is 0, it is on the point of turning around and has therefore reached maximum displacement from the equilibrium position. This displacement is called the amplitude, $a$. In other words:

$v = 0 \iff x = a$

$\begin{aligned} \therefore \dfrac{1}{2} \omega^2 a^2 = C & \ \Rightarrow \ \dfrac{1}{2} v^2 = \dfrac{1}{2} \omega^2 (a^2 - x^2) \\ & \ \Rightarrow v^2 = \omega^2 (a^2 - x^2) \\ & \ \Rightarrow v = \pm \omega \sqrt{a^2 - x^2} \end{aligned}$

Since velocity is the rate of change of displacement.

$\dfrac{dx}{dt} = \pm \omega \sqrt{a^2 - x^2}$

This integration can be performed by the separation of the variables:

$\displaystyle \int \dfrac{1}{\sqrt{a^2 - x^2}}\ dx = \pm \omega \int dt$

A substitution cracks the integral on the LHS and the RHS is easy as $\pi$

$Let \ x = a \sin \theta \Rightarrow \dfrac{dx}{d\theta} = a \cos \theta$

$\displaystyle \int \dfrac{a \sin \theta}{\sqrt{a^2 (1-\sin^2 \theta)}}\ d\theta = \pm \omega \int dt$

$\displaystyle \int d\theta = \pm \omega \int dt$

$\theta = \pm \omega t + C$

From the substitution, $\theta = \arcsin \left( \dfrac{x}{a} \right)$

$\arcsin \left( \dfrac{x}{a} \right) = \pm \omega t + C$

$\dfrac{x}{a} = \sin (\omega t + C)$

$x = a \sin (\omega t + C)$

Oh, how is ωt related to theta?

theta = ωt
It's the angular equivalent of s = vt

oh ok, x=Asin(ωt) how do i find the max velocity from this?
And then max acceleration.

thanks

There is a bunch of formulas for SHM which give such things as max velocity and max acceleration.
If you differentiate that expression wrt time you have velocity.
The maximum velocity occurs when the cosine = 1
So you should be able to work out what that becomes.
The max acceleration comes from the defining equation and occurs at maximum displacement where the restoring force is maximum. When x = amplitude.

if a is max when x=A then why isn't it this:

$amax = -\omega^2 A$

What do you mean? Isn't this?

The defining equation is
$a = -\omega^2 x$
So the maximum value of a occurs when x is equal to the amplitude. The maximum value of x.

So what are you asking?

$amax = \omega^2 x$

Why did they miss out the -?
Is it so that acceleration points towards the equilibrium position?

Who missed it out? Where?
In all your posts before #14 there has been a minus sign.