Answer
$.54 Hz; .22 m; -6^{\circ}$
Work Step by Step
We find the velocity of the two blocks after the collision:
$v = \frac{.8\times1.7}{2}=.68 \ m/s$
We first find the amplitude increase. We know that the new block will add energy to the system. Thus, we find:
$\frac{1}{2}kx_0^2 +\frac{1}{2}mv^2= \frac{1}{2}kx_f^2$
$\frac{1}{2}(23)(.1)^2 +\frac{1}{2}(2)(.68)^2= \frac{1}{2}(23)x_f^2$
Thus, the amplitude is:
$x_f=22.4 \ cm = .22 \ m$
We now find the new frequency:
$f=\frac{2\pi}{\sqrt{\frac{k}{m+M}}}$
$f=\frac{\sqrt{\frac{23}{2}}}{2\pi}=.54\ Hz$
We now find the phase constant:
$\phi = tan^{-1}(\frac{v}{2\pi fA_0})-90\sqrt{\frac{M}{m+M}}$
$\phi = tan^{-1}(\frac{.68}{2\pi (.54)(.1)})-90\sqrt{\frac{1.2}{2}}=-6^{\circ}$
