Finding the Phase Constant

Solution:

$v\left(t\right)=-A\omega\sin\left(\omega t+\varphi\right)$

$v_{\max}=A\omega\ =0.4\ cm$ from the plot!

$T=\pi$ (for example from $t=-\frac{1}{6}\pi$ to $t=\frac{5}{6}\pi$)

$\Rightarrow\omega=\frac{2\pi}{T}=2\ rad/s\Rightarrow A=0.2\ cm$

The last thing we need to find is the phase constant:

$v\left(0\right)=-A\omega\sin\left(\omega\left(0\right)+\varphi\right)=-0.35$

$\Rightarrow\sin\left(\varphi\right)=\frac{0.35}{0.4}=\frac{7}{8}$

This gives us two possible answers (61 degrees or 119 degrees). Which is it?

To figure that out, we will look at the acceleration function:

$a\left(t\right)=-A\omega^2\cos\left(\omega t+\varphi\right)$

$\Rightarrow a\left(0\right)=-A\omega^2\cos\left(\omega\left(0\right)+\varphi\right)$

$\rightarrow\ \ \ \ \ \ \ \ -\cos\left(\varphi\right)<0\ \ \ \ \ \ \rightarrow\ \ \ \ \ \cos\left(\varphi\right)>0$

We know the acceleration at t=0 is negative from the graph (slope of the tangent tine to $v\left(t\right)$ at $t=0$ is negative!)

$\Rightarrow\cos\left(\varphi\right)>0$

Only the 61 degree answer satisfies this condition.

So, we can now plot $x\left(t\right)=A\cos\left(\omega t+\varphi\right)=\boxed{0.2\cos\left(2t+1.06\right)cm}$

Note: We need to convert the 61 degrees to 1.06 radians because the "2t" is also in radians. (The 2 is in radians per second.)

The plot below is in seconds on the horizontal axis.