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=-\frac12\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(c_2)/(c_1)`.

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


`x'(t)=\frac12\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\\\\-\frac12C_1+10C_2=1\end{cases}\implies C_1=1,C_2=\frac3{20}\implies R=(√(409))/(20),\delta=\arctan\frac3{20}`

So the particular solution is


`x(t)=(√(409))/(20)\cos\left(10t-\arctan\frac3{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.