I am trying to find the new amplitude of an oscillating mass-spring system after applying an impulse forcing function on it. The equation for the position of the mass (end of the spring) is given by $$x''+16x=f(t)$$ where $f(t)$ is defined as: $$ f(t) = \begin{cases} 0, & t<4 \\ k, & 4\leq x<4.1 \\ 0, & 4.1\leq t \end{cases} $$ and the initial conditions are $x(0)=-4$ and $x'(0)=-16$.
If you solve for $x$ you get:
$$X\left(t\right)=-4\sin\left(4t\right)-4\cos\left(4t\right)+\frac{k}{16}\left(u\left(t-4\right)\left(1-\cos\left(4t-16\right)\right)-u\left(t-4.1\right)\left(1-\cos\left(4\left(t-4.1\right)\right)\right)\right)$$
where $u(x)$ is the unit step function. My question is how to find the amplitude of the oscillations of $X(t)$ when $t>4.1$ in terms of $k$.
My attempt:
$X(t)$ can be written as the sum of the two functions: $$M\left(t\right)=-4\sin\left(4t\right)-4\cos\left(4t\right)$$ $$G(t)=\frac{k}{16}\left(u\left(t-4\right)\left(1-\cos\left(4t-16\right)\right)-u\left(t-4.1\right)\left(1-\cos\left(4\left(t-4.1\right)\right)\right)\right)$$ where $$X(t)=M(t)+G(t)$$ however, since we only care about oscillations at $t>4.1$, we can rewrite $G(t)$ as: $$G\left(t\right)=\frac{k}{16}\left(\cos\left(4t-16.4\right)-\cos\left(4t-16\right)\right)$$ To find the extrema of $X(t)$, you take the derivative and set it equal to $0$: $$X'(t)=M'(t)+G'(t)=0$$ Calculating the derivative of each, we get that: $$M'(t) = -16 (\cos(4 t) - \sin(4t))$$ $$G'(t) = \frac{k}{4}(\sin(16.4 - 4 t) - \sin(16 - 4 t))$$ So: $$ \frac{k}{4}(\sin(16.4 - 4 t) - \sin(16 - 4 t))-16 (\cos(4 t) - \sin(4t))=0 $$ This is where I got stuck since I do not know how to solve this equation for $t$ and could not find anything online to help.
Note: After finishing to write up this question I realized that there is no need to write $X(t)$ as a sum of two other functions for this problem. The reason I did that initially is because I thought about the maximum of $X(t)$ as the point where $M'(t)=-G'(t)$ since I had each of those functions graphed on Desmos.
I'm not going to check if your solution is correct.
Derivatives are not necessary here. You should know that linear combinations of sinusoids with the same frequency, is itself another sinusoid.
For $t > 4.1$, the solution simplifies
\begin{align} x(t>4.1) &= -4\sin (4t) - 4\cos(4t) + \frac{k}{16}\bigg[-\cos\big(4(t-4)\big) + \cos\big(4(t-4.1)\big)\bigg] \\ &= -4\sin (4t) - 4\cos(4t) \\ &\quad + \frac{k}{16}\bigg[-\cos(16)\cos(4t)-\sin(16)\sin(4t) + \cos(16.4)\cos(4t) + \sin(16.4)\sin(4t)\bigg] \\ &= A\sin(4t) + B\cos(4t) \end{align}
where (after combining coefficients) \begin{align} A &= -4 + \frac{k}{16}\bigg[\cos(16.4) - \cos(16)\bigg] \\ B &= -4 + \frac{k}{16}\bigg[\sin(16.4) - \sin(16)\bigg] \end{align}
The amplitude of this is simply $$ \max_{t > 4.1} |x(t)| = \sqrt{A^2 + B^2} $$