# Simple Harmonic Motion given velocity and acceleration

I am trying to understand how to relate velocity and acceleration of an object to it's amplitude, period, and frequency given only the following:

An object of mass $m=20\ \rm kg$ moves with SHM along the $x$-axis. At time $t=0$ it is located at a distance of 4m away from the origin which is located at $x=0$, and has a velocity $v=15\ \rm ms^{-1}$ and acceleration, $a=10\ \rm ms^{-2}$ which is directed towards the origin.

I tried setting it up using the EOM of $x(t)=A\cos(\omega{t}-\phi)$ because it is 1D motion toward the axis. However, this would mean that $v(t)=\frac{\mathrm{d}x(t)}{\mathrm{d}{t}}=-A\omega\sin(\omega{t})$, which gives $v(0)=0$, and is not equal to the given $15\ \rm m/s$. The same issue arises with acceleration if I use $x(t)=A\sin(\omega{t}+\phi)$. Therefore, if my initial conditions are not $v(0)=0=a(0)$, what would the EOM be?

• When you differentiate, you should keep the $\pm\phi$ in the $\sin()$ or $\cos()$ functions. (I wrote $\pm$ because you have it written both ways in your question.) – Mike Bell Feb 6 '16 at 21:44
• Ok, but in this case, what would the phase be? – NotSoSN Feb 6 '16 at 21:53
• The phase is something you have to solve for, based on the initial conditions. You have three unknowns ($\omega$, $A$, and $\phi$), and you have three conditions/equations to use to solve for those unknowns. So find expressions for $a(t)$, $v(t)$, and $x(t)$, and then make use of the initial conditions. – Mike Bell Feb 6 '16 at 22:07

You need to find $\omega$ first - think about the definition of shm.
A very useful relationship when trying to solve this sort of problem is $v^2=\omega^2(A^2-x^2)$. Having used that relationship you can sort out the form of the equation for $x$ and the phase.