Thread: [SOLVED] Simple Harmonic Motion

How do you prove that a function shows that something is in SHM if the equation isn't in the form:

x = Acos (kt + b)

Is a graph would be acceptable or is there a mathematical way to do this?

Originally Posted by Evales

How do you prove that a function shows that something is in SHM if the equation isn't in the form:

x = Acos (kt + b)

Is a graph would be acceptable or is there a mathematical way to do this?

Post the function.

x = sin t + [(3)^(1/2)]cos t

Originally Posted by Evales

How do you prove that a function shows that something is in SHM if the equation isn't in the form:

x = Acos (kt + b)

Is a graph would be acceptable or is there a mathematical way to do this?

(Edited version)

The equation you have written is one way to define harmonic motion. So if your trial function cannot be put into this form, then it does not represent a SHO.

On the other hand you can define simple harmonic motion as $\displaystyle x"(t) = -Ax(t)$. By taking small displacements from equilibrium for many physical systems you will find it is a SHO. (Usually you need to do a Taylor expansion to prove this. The linear coefficients in the series needs to be 0 and the quadratic coefficient needs to be negative. The condition on the linear term is automatic around an equilibrium point, so many systems have this kind of behavior.)

For your function
$\displaystyle sin(t) + A~cos(t)$
you can easily see that it is the (linear) sum of two SHO functions, so it is also a harmonic oscillator.

-Dan

Originally Posted by topsquark

For your function
$\displaystyle sin(t) + A~cos(t)$
you can easily see that it cannot be put into $\displaystyle cos(kt + b)$ form.

You might easily be able to see that it cannot be put in that form, but I can easily see that it can .

RonL

Originally Posted by CaptainBlack

You might easily be able to see that it cannot be put in that form, but I can easily see that it can .

RonL

Any way of enlightening me?

Originally Posted by CaptainBlack

You might easily be able to see that it cannot be put in that form, but I can easily see that it can .

RonL

Yeah, I was just coming back to fix that.

-Dan

Originally Posted by Evales

Any way of enlightening me?

$\displaystyle x = sin(t) + A~cos(t)$

$\displaystyle x' = cos(t) - A~sin(t)$

$\displaystyle x" = -sin(t) - A~cos(t) = -x$

So this is a harmonic oscillator.

Or
$\displaystyle sin(t) + A~cos(t) = \sqrt{1 + A^2}~sin(t + \phi)$
(where $\displaystyle \phi$ is a constant) which is a sine function and hence a harmonic oscillator.

(This is from the more general:
$\displaystyle a~sin(t) + b~cos(t) = \sqrt{a^2 + b^2}~sin(x + \phi)$
where
$\displaystyle \phi = tan^{-1} \left ( \frac{b}{a} \right )$
for $\displaystyle a \geq 0$
and
$\displaystyle \phi = tan^{-1} \left ( \frac{b}{a} \right ) + \pi$
for $\displaystyle a < 0$)

-Dan