If,
$\dfrac{d^2x}{dt^2} = -16x$
When $t= 0, v = 10$ and $x = 0$, where $x$ is displacement and $v$ is velocity.
How would I obtain an expression for the velocity and displacement at time t?
I am used to working with $dv/dt$, $vdv/dt$ and $d/dx[1/2v^2]$ and I'm not sure how to go about this question.
My solution so far:
$\dfrac{d^2x}{dt^2} = -16x$
$\dfrac{d^2x}{-16x} = dt^2$
This is where I get stuck, according to wolfram alpha the final solution will be trigonometric.
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$\begingroup$Since there seem to be some confusion, I write some more details of a solution.
Your differential equation can be written $$ x''(t)+16x(t)=0. $$ The characteristic polynomial is $$ r^2+16=0, $$ with solutions $r=\pm 4i$. Therefore, by the theorem on linear second order differential equations with constant coefficients, $$ x(t)=A\cos 4t+B\sin 4t, $$ for some constants $A$ and $B$. Now use the fact that $x(0)=0$ and $x'(0)=10$ to determine $A$ and $B$.
Alternative way of solving the differential equation
Multiply by $x'$, to get $$ x''(t)x'(t)+16x'(t)x(t)=0. $$ Now $$ x''(t)x'(t)=\frac{1}{2}\frac{d}{dt}(x'(t))^2\quad\text{and}\quad x'(t)x(t)=\frac{1}{2}\frac{d}{dt}(x(t))^2 $$ and we conclude that $$ \frac{d}{dt}((x'(t))^2+16x(t)^2)=0. $$ Integrating from $0$ to $t$, we find that $$ (x'(t))^2+16x(t)^2=(x'(0))^2+16x(0)^2=100. $$ Thus, $$ x'(t)=\pm\sqrt{100-16x(t)^2}, $$ where the plus has to be taken, since $x'(0)=10$. This is a separable differential equation, $$ \frac{x'(t)}{\sqrt{100-16x(t)^2}}=1. $$ Integrating from $0$ to $t$, $$ \frac{1}{4}\arcsin(2x(t)/5)=t. $$ Moving things around, we get $$ x(t)=\frac{5}{2}\sin(4t). $$
$\endgroup$ 6 $\begingroup$Philip wrote :
$\dfrac{d^2x}{dt^2} = -16x$
$\dfrac{d^2x}{-16x} = dt^2$
This is a complete nonsens.
It is more important for Philip to understand why this is a nonsens, than to solve the equation.
That is the raison why I answer to him without even talking about the method of solving.
$\frac{d^2 x}{dt^2}$ doesn't mean that $d^2x$ is devided by $dt^2$. This is a conventional way to say that the function $x(t)$ is differentiated two times successively.
So literally $\frac{d^2x}{dt^2} = -16x$ means : the second derivative of the function $x(t)$ is equal to minus 16 times the function $x(t)$ itself. There is no $d^2x$ nor $dt^2$ in this sentense, so you are not allowed to mutiply or divide by one of these terms which have no meaning when taken independently.
$\endgroup$ 2 $\begingroup$Here is a way of doing this (write $\frac {dx}{dt}=x'; \frac {d^2x}{dt^2}=x''$) by separating variables. I don't know whether you will be familiar with all the ideas, but they may well become familiar.
With the equation $\frac {dx}{dt}=ax$ separating the variables gives $\int\frac {dx}x=\int a dt$ so that $\ln x=at+c$ and raising to the power $e$ with $C=e^c$ gives $x=Ce^{at}$.
Now, with $x''=-a^2x$ write $y=x'+aix$ and $z=x'-aix$ where $i$ is a square root of $-1$.
Then $$y'=x''+aix'=-a^2x+aix'=ai(aix+x')=aiy$$ so we have $y=Ce^{ait}$ and in a similar way $z=De^{-ait}$
Now compute $y'-z'=2aix=Ce^{ait}+De^{-ait}$ and finally dividing through by $2ai$ we have $x=Ae^{ait}+Be^{-ait}$
Then we can use $e^{\pm ait}=\cos at \pm i \sin at$ and to transform this into the form $x=E\cos at+ F\sin at$
For the equation $$x''+ax'+bx=0$$ let $\alpha$ and $\beta$ be the roots of $u^2-au+b=0$ so that $a=-(\alpha+\beta)$ and $b=\alpha\beta$ then you will find that $y=x'-\alpha x$ and $z=x'-\beta x$ allow you to separate variables e.g. $$0=x''-(\alpha+\beta)x'+\alpha\beta x=(x''-\alpha x')-\beta (x'-\alpha x)=y'-\beta y$$ so that $y=Ce^{\beta t}$ and similarly $z=De^{\alpha t}$ then $y-z=(\beta-\alpha)x=Ce^{\alpha t}+De^{\beta t}$ and if $\beta \neq \alpha$ then $x=Ae^{\alpha t}+Be^{\beta t}$
It is this form which you will find yourself using if you do much more work of second order differential equations of this kind - you don't have to do all the workings every time, just solve the quadratic and plug in the roots.
$\endgroup$ $\begingroup$Hint:
As noted in the comments there is a general method to solve second order linear differential equations, that you can find on any book about differential equations. But this case is very simple.
I'm pretty sure that you know that a function as: $ f(t)=\sin (at)$ has first derivative $f'(t)=a \cos (at)$ and second derivative: $ f''(t)= -a^2\sin (at)=-a^2 f(t)$. And we have the same for a function $g(t)=\cos(at)$, that gives:$ g''(t)= -a^2\cos (at)=-a^2 g(t)$.
Since you have a linear equation a solution is linear combination of these particular solutions: $ x(t)= C_1 f(t)+C_2 g(t)$
Can you elaborate from this and find the two cosntants from the initial conditions?
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