Question

Only the p=0.5 is correct how about the others?

Answer 1

One possible problem with your solution is that it contains $\sqrt t$ 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 $s(t)$ satisfies the second-order ODE

$4x''+4x'+401x=0$

(assuming there are no other external forces acting on the mass). The characteristic equation for this ODE is

$4r^2+4r+401=0\implies r=-\dfrac12\pm10i$

which means the general solution to this ODE is

$x(t)=\bigg(C_1\cos(10t)+C_2\sin(10t)\bigg)e^{-t/2}$

The angle difference identity for cosine allows you to condense the trigonometric part of the solution to

$C_1\cos(10t)+C_2\sin(10t)=R\cos(10t-\delta)$

where $C_1=R\cos\delta$ and $C_2=R\sin\delta$, leaving you with

$x(t)=R\cos(10t-\delta)e^{-t/2}$

These unknown constants can be found explicitly, as $R=\sqrt{{C_1}^2+{C_2}^2}$ and $\delta=\arctan\dfrac{c_2}{c_1}$.

Given that $x(0)=1$ and $v(0)=x'(0)=7$, and the solution's first derivative is

$x'(t)=\dfrac12\bigg((20C_2-C_1)\cos(10t)-(20C_1+C_2)\sin(10t)\bigg)e^{-t/2}$

you have the following system of equations needed to find $C_1,C_2$, and from there the corresponding values of $R$ and $\delta$.

$\begin{cases}C_1=1\\\\-\dfrac12C_1+10C_2=1\end{cases}\implies C_1=1,C_2=\dfrac3{20}\implies R=\dfrac{\sqrt{409}}{20},\delta=\arctan\dfrac3{20}$

So the particular solution is

$x(t)=\dfrac{\sqrt{409}}{20}\cos\left(10t-\arctan\dfrac3{20}\right)e^{-t/2}$
$x(t)\approx1.0112\cos(10t-0.1489)e^{-0.5t}$

In terms of what you should submit, you would use

$C_1=1.0112$
$\omega_1=10$
$\alpha_1=0.1489$
$p=0.5$

or rely on the exact forms in case rounded answers are not accepted.