1. ## Sinusodial distance formula

Well most of you know that the formula of distance in an oscillating (harmonic) movement is x = r.coswt. I know how to get at this formula using the circular movement. But I gotta reach it by using the acceleration.

K.................................O...........obje ct............. L
|------------------------|-----------o------------|
...................................|_________|
...........................................x

x = distance from the center
k = N/x (like the k of a spring)
m = mass
a = acceleration

At any point, F = x*k
and F = m*a so
a = x*k/m

But I cant figure out how to find the sinusodial distance formula from here.

2. ## Calculus needed

First, you need to remember that the force, and thus acceleration, is in the opposite direction of the displacement x, so we have $F=-k \cdot x$ and $a=-\frac{k}{m}x$.

From here, we need calculus:

Remember that $a = \frac{dv}{dt} = \frac{d^{2}x}{dt^2}$

so we have $\frac{d^{2}x}{dt^2} = -\frac{k}{m}x$

Solving this differential equation gives us the sinusoidial solution.

(Hint: define $\omega_0 = \sqrt{\frac{k}{m}}$)

--Kevin C.

3. Ok. That equation gave me,
$x = V_0 \text{Cos}\left(t\sqrt{\frac{k}{m}}\right) + x_0\text{Sin}\left(t\sqrt{\frac{k}{m}}\right)
$

As $x_0 = 0$ on the starting point, I cancel the sine function.

So, $x = V_0 \text{Cos}\left(t\sqrt{\frac{k}{m}}\right))
$

$\omega_0 = \sqrt{\frac{k}{m}}$
$V_0 = r\omega_0$
$x = \omega r Cos(\omega t)$

I think I did something wrong ..

4. ## Constants are off

From the differential equation $\frac{d^{2}x}{dt^2} = -\frac{k}{m}x$, which we use $\omega \equiv \sqrt{\frac{k}{m}}$ to rewrite as $\frac{d^{2}x}{dt^2} = -\omega^{2}x$,
you should get a general solution of the form $x(t)=A\cos (\omega t) + B\sin (\omega t)$.

Putting in t=0, and using $\cos (0) = 1$ and $\sin (0) = 0$, we get $x_0 = x(0) = A \cdot 1 + B \cdot 0 = A$.
Similarly, $v(t)=x'(t)=-A\omega \sin (\omega t) + B\omega \cos (\omega t)$, and $v_0=v(0)=-A\omega \cdot 0 + B\omega \cdot 1 = B\omega$, and $B=\frac{v_0}{\omega}$
Thus you should have $x(t)=x_0 \cos (\omega t) + \frac{v_0}{\omega} \sin (\omega t)$, not $x(t)=v_0 \cos (\omega t) + x_0 \sin (\omega t)$.

--Kevin C.

5. Ok, I just thought it was an ordinary integration constant (like in linear motion). Thanks for helping out.