# If i know velocity, then i know acceleration right?

1. Jan 11, 2005

### nemzy

For velocity at a certain particle in a simple harmonic oscillation is the following:

v= w (anguluar frequency) times the squareroot of A^2-x^2

but if i wanted to find the acceleration at a certain particle in a simple harmonic oscillation, i can somehow derive a formula from the above equation right? But how would u derive it?

I know that that V= dx/dt ..and a= (d^2)x/dt^2

ugh, i forgot my calculus, anyone clear it up for me?

thanks

2. Jan 11, 2005

### vincentchan

for simple hamonic motion
$$x = A \sin{( wt-\theta)}$$
I have no idea what is your equation looks like

v= w (anguluar frequency) times the squareroot of A^2-x^2

put it in Latex pls.....

3. Jan 11, 2005

### dextercioby

The equations for linear harmonic oscillator are something like that
$$x(t)=A\sin(\omega t+\phi)$$
$$v_{x}(t)=A\omega\cos(\omega t+\phi)$$
$$a_{x}(t)=-A\omega^{2}\sin(\omega t+\phi)$$

Then decide what are the initial conditions.And u can determine the 2 unknowns;

Daniel.

4. Jan 11, 2005

### vincentchan

$$v= w \sqrt{A^2-x^2}$$ is that what you trying to say???

5. Jan 11, 2005

### vincentchan

OKAY, i dunno where did you get this equation.... this is right but ppl usually don't write it this way, if you wanna find a(t) from your equation, substitude v=dx/dt, you have
$$dt=dx/ (w \sqrt{A^2-x^2})$$
integrate both side and you will have$$x(t) = Asin(wt+\theta)$$
after you have x(t), everything should be easy

6. Jan 11, 2005

### nemzy

yes,

$$v= w \sqrt{A^2-x^2}$$

is what i am trying to say, so how can u derive a formula from the above equation for acceleration? that is my question

7. Jan 11, 2005

### vincentchan

did i just tell you.. do the integral and find x(t)
after you have x(t), v(t)= dx/dt, a(t) = dv/dt

8. Jan 11, 2005

### dextercioby

The formula is deduced from applying the law of energy conservation.Once u integrate this ODE (using the method prescribed above),you need to diff.2 times wrt to time to find the acceleration.

Daniel.