Simple harmonic motion equation

[Taylan]

Homework Statement

Calculate the harmonic motion equation for the following case
A=0.1m, t=0s x=0.05m, v(t=0)>0 a(t=0)= -0.8m/s^2

Homework Equations

x(t)= +/-Acos/sin ( (2pi/T)/*t)

The Attempt at a Solution

[/B]
A is given to be 0.1 so I simply place it into the equation. Now I have to decide whether the graph is sin or cos and the sign. Since the velocity is positive at t=0, the sign must be +. I am confused about whether it is sin or cos because the graph when I sketch it looks like a kind of transformation starting from t=0 x=0.05m and going upwards. I am also not sure about how to find T. is there an equation I am missing?

 

[PeroK]

Homework Statement

Calculate the harmonic motion equation for the following case
A=0.1m, t=0s x=0.05m, v(t=0)>0 a(t=0)= -0.8m/s^2

Homework Equations

x(t)= +/-Acos/sin ( (2pi/T)/*t)

The Attempt at a Solution

[/B]
A is given to be 0.1 so I simply place it into the equation. Now I have to decide whether the graph is sin or cos and the sign. Since the velocity is positive at t=0, the sign must be +. I am confused about whether it is sin or cos because the graph when I sketch it looks like a kind of transformation starting from t=0 x=0.05m and going upwards. I am also not sure about how to find T. is there an equation I am missing?

If the motion at $t=0$ is neither at $x=0$ nor $x = \pm A$, then the motion needs a phase angle.

 

[Taylan]

If the motion at $t=0$ is neither at $x=0$ nor $x = \pm A$, then the motion needs a phase angle.

Thank you! So:

0.1cos[(2pi/T)*0 + phi) = 0.05

phi= pi/3

What I end up with is:

x(t) = 0.1cos[(2pi/T)*t+pi/3)

Can please you give me further tips to find out the value of T?

 

[Delta2]

if you find $x(t)$ then the equation for acceleration is $a(t)=-\omega^2x(t)$ and because it is given $a(t=0)=-0.8$ you ll have the following equation
$a(0)=-0.8=-\omega^2x(0)$ with only unknown the $\omega=\frac{2\pi}{T}$