One possible problem with your solution is that it contains [tex]\sqrt t[/tex] in the argument of cosine, when it should be a linear term. Aside from that, the best way to track down a mistake is to start from the beginning:
The mass's position function [tex]s(t)[/tex] satisfies the second-order ODE
[tex]4x''+4x'+401x=0[/tex]
(assuming there are no other external forces acting on the mass). The characteristic equation for this ODE is
[tex]4r^2+4r+401=0\implies r=-\dfrac12\pm10i[/tex]
which means the general solution to this ODE is
[tex]x(t)=\bigg(C_1\cos(10t)+C_2\sin(10t)\bigg)e^{-t/2}[/tex]
The angle difference identity for cosine allows you to condense the trigonometric part of the solution to
[tex]C_1\cos(10t)+C_2\sin(10t)=R\cos(10t-\delta)[/tex]
where [tex]C_1=R\cos\delta[/tex] and [tex]C_2=R\sin\delta[/tex], leaving you with
[tex]x(t)=R\cos(10t-\delta)e^{-t/2}[/tex]
These unknown constants can be found explicitly, as [tex]R=\sqrt{{C_1}^2+{C_2}^2}[/tex] and [tex]\delta=\arctan\dfrac{c_2}{c_1}[/tex].
Given that [tex]x(0)=1[/tex] and [tex]v(0)=x'(0)=7[/tex], and the solution's first derivative is
[tex]x'(t)=\dfrac12\bigg((20C_2-C_1)\cos(10t)-(20C_1+C_2)\sin(10t)\bigg)e^{-t/2}[/tex]
you have the following system of equations needed to find [tex]C_1,C_2[/tex], and from there the corresponding values of [tex]R[/tex] and [tex]\delta[/tex].
[tex]\begin{cases}C_1=1\\\\-\dfrac12C_1+10C_2=1\end{cases}\implies C_1=1,C_2=\dfrac3{20}\implies R=\dfrac{\sqrt{409}}{20},\delta=\arctan\dfrac3{20}[/tex]
So the particular solution is
[tex]x(t)=\dfrac{\sqrt{409}}{20}\cos\left(10t-\arctan\dfrac3{20}\right)e^{-t/2}[/tex]
[tex]x(t)\approx1.0112\cos(10t-0.1489)e^{-0.5t}[/tex]
In terms of what you should submit, you would use
[tex]C_1=1.0112[/tex]
[tex]\omega_1=10[/tex]
[tex]\alpha_1=0.1489[/tex]
[tex]p=0.5[/tex]
or rely on the exact forms in case rounded answers are not accepted.
