I don't think that anyone has ever calculated this (correct me if I'm wrong on this), so I want to ask professionals if it is right or wrong?
we know that the force that controls arrows in the "arrow and bow" weapon is "spring force"
what I have done is calculating how much long you must pull the arrow (let's say $(n)$ meters long) to make the initial velocity for the arrow equal some constant $(vi)$
We know that $F$(spring)=-$kx$ , and $F=ma$
$-kx=ma$
$-kx=m*dv/dt$
$-kx=m*d^2x/dt^2$
solving this differential equation gives us
$x(t) =c1*cos(\sqrt{(k/m)}*t) + c2*sin(\sqrt{(k/m)}*t)$
t is zero the second you release the arrow, and when you want to release the arrow x will be $(-n)$, (if we pull the arrow to the left and assume that when $n=0 , x=0 )$
and if we put $x=-n$ and $t=0$ we will get $c1=-n$
therefore
$x(t) =-n*cos(\sqrt{(k/m)}*t) + c2*sin(\sqrt{(k/m)}*t)$
and the second you release the arrow $(t=0)$ the velovity will be $0 (x'(0)=0)$
$x'(t) = \sqrt{(k/m)}*n*sin(\sqrt{(k/m)}*t) + \sqrt{(k/m)}*c2*cos(\sqrt{(k/m)}*t)$
and if we put $x'=0$ and $t=0$ we will get $c2=0$
therefore the equation will be
$x(t)=-n*cos(\sqrt{(k/m)}*t)$
where n is how much long you pull the arrow , and m is the mass of the arrow , and k is the spring constant , and t is time
now we want to know what is the initial velocity that the arrow will travel with when it get out of the bow, and for that we want to know the velocity at the second when the force stops affecting on the arrow, because the acceleration will become $0$ and the velocity will become a constant velocity, and this all happens when $x(t)=0 $
$x(t)=0$
$-n*cos(\sqrt{(k/m)}*t)=0$
Arrows propelled by spring force of bow
$cos(\sqrt{(k/m)}*t)=0$
(and that happens just for one time)
therefore $ \sqrt{(k/m)}*t = \pi/2$
hence, $ t=(\sqrt{(m/k)}*\pi)/2$
that is the time when the tail of the arrow reaches the point $x=0$ and its velocity at that time will be its initial velocity
so $vi = x'((\sqrt{(m/k)}*\pi)/2) = \sqrt{(k/m)}*n * sin(\sqrt{(k/m)}* (\sqrt{(m/k)}*\pi)/2)$
$ vi=\sqrt{(k/m)}* n *sin(\pi/2)$
therefore $vi = n*\sqrt{(k/m)}$
so the initial velocity the arrow will travel with is how much long you pull the arrow times the square root of (the spring constant over the mass of the arrow).
Do these calculations seem to be reasonable as regards the analysis of the problem ?
Best Answer
If you are going to treat a bow/arrow as a Hookean spring - mass system (probably a highly inaccurate representation, actually) then finding the initial velocity $v_i$ (that is, to be clear, the velocity with which the arrow leaves the bow when released) can be done in a much simpler way. No equation of motion of the arrow is needed and we simply use Conservation of Energy.
As the displacement of the spring is $x$ it can be shown very easily that the spring now contains potential energy $U$ (Work - Energy Theorem $ ^*$), determined by:
$$U=\frac12 kx^2$$
As the arrow is released that spring potential energy is converted to kinetic energy of the arrow, $K$:
$$K=\frac12 mv^2$$
Where $m$ and $v$ are the mass and velocity of the arrow respectively.
For $x=0$ (we assume no friction or drag losses and massless bow, of course):
$$U=K \implies \frac12 kx^2=\frac12 mv_i^2$$ $$\implies v_i=\sqrt{\frac{k}{m}}x$$
Your derivation is basically correct but a bit 'going from Paris to London via Oslo'!
$ ^*$
The work $W$ done on displacing the arrow against the restoring force of the bow is also the potential energy $U$ stored in it:
$$U=W=\int_0^xF(x)dx=\int_0^xkxdx=\frac12 kx^2$$