Spring Differential Equation

[Drakkith]

Homework Statement

Consider the illustration of 3 springs:

In A, we hang a very light spring and pan from a hinge. The pan and spring are so light, we can neglect any stretching of the original length $l_{0}$. In B we add a weight $mg$ which force is balanced by $kl$ (Hooke's Law; the spring tension is proportional to the displacement $l$); i.e. $mg=kl$.
In C, we displace the pan by a further $y(0) = y_{0}$ and give the pan an initial velocity $v_{0} = \frac{dy}{dt}|_{t=0}$.

Let $y(t),v(t)=\frac{dy}{dt}$ be the displacement and velocity of the pan. $y(t)$ satisfies the equation (Newton's 2nd Law): $m\frac{d^2y}{dt^2}=mg-k(y+l)-B\frac{dy}{dt}$
or, since $mg = kl$, and dividing across by $m$,

$(1)$ $\frac{d^2}{dt^2}+\frac{B}{m}\frac{dy}{dt}+\frac{k}{m}y=0$

Given $\frac{B}{m} = 4(sec)^{-1}$ and $\frac{k}{m} = (a.) 3, (b.) 4, (c.) 7$,
with $y(0) = 1, \frac{dy}{dt}(0) = 0$, solve $(1)$ for the three cases, $a, b, c$.
Write the solution in each case for $y(t), v(t)$.

Homework Equations

$y(0) = 1$
$\frac{dy}{dt}(0) = 0$

The Attempt at a Solution

I think I've solved two of the cases they wanted me to solve, but I'm stuck on this part:

Case $(a.)$
As $t→∞$, the vector $\left(\frac{y(t)}{v(t)}\right)$ approaches $\left(\frac{0}{0}\right)$ along what direction?

Could someone explain what this means? What does 'approaches along what direction' mean? The three possible answers are $\left(\frac{1}{1}\right), \left(\frac{1}{-1}\right), \left(\frac{-1}{3}\right)$. (They aren't fractions, I just don't know how to put a vertical vector in a post using LaTex)

 

[Ray Vickson]

Homework Statement

Consider the illustration of 3 springs:
View attachment 113883

In A, we hang a very light spring and pan from a hinge. The pan and spring are so light, we can neglect any stretching of the original length $l_{0}$. In B we add a weight $mg$ which force is balanced by $kl$ (Hooke's Law; the spring tension is proportional to the displacement $l$); i.e. $mg=kl$.
In C, we displace the pan by a further $y(0) = y_{0}$ and give the pan an initial velocity $v_{0} = \frac{dy}{dt}|_{t=0}$.

Let $y(t),v(t)=\frac{dy}{dt}$ be the displacement and velocity of the pan. $y(t)$ satisfies the equation (Newton's 2nd Law): $m\frac{d^2y}{dt^2}=mg-k(y+l)-B\frac{dy}{dt}$
or, since $mg = kl$, and dividing across by $m$,

$(1)$ $\frac{d^2}{dt^2}+\frac{B}{m}\frac{dy}{dt}+\frac{k}{m}y=0$

Given $\frac{B}{m} = 4(sec)^{-1}$ and $\frac{k}{m} = (a.) 3, (b.) 4, (c.) 7$,
with $y(0) = 1, \frac{dy}{dt}(0) = 0$, solve $(1)$ for the three cases, $a, b, c$.
Write the solution in each case for $y(t), v(t)$.

Homework Equations

$y(0) = 1$
$\frac{dy}{dt}(0) = 0$

The Attempt at a Solution

I think I've solved two of the cases they wanted me to solve, but I'm stuck on this part:

Case $(a.)$
As $t→∞$, the vector $\left(\frac{y(t)}{v(t)}\right)$ approaches $\left(\frac{0}{0}\right)$ along what direction?

Could someone explain what this means? What does 'approaches along what direction' mean? The three possible answers are $\left(\frac{1}{1}\right), \left(\frac{1}{-1}\right), \left(\frac{-1}{3}\right)$. (They aren't fractions, I just don't know how to put a vertical vector in a post using LaTex)

$$ \pmatrix{1\\-1} \; \text{or} \; \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$
These work for any number of dimensions:
$$\pmatrix{a\\b\\c} \; \text{or} \; \begin{bmatrix}a\\b\\c \end{bmatrix}.$$
Right-click on the formula and select "display math as tex..." to see the commands used. Alternatively, if you have only two components you can "cheat" and use the binomial coefficient, like this:
$${1 \choose -1}. $$