0
$\begingroup$

I have this problem where I am trying to calculate $d(t)$ and $v(t)$ of a mass m on a spring, dropped from a displacement $A$, without using anything else than Hooke's law and energy calculations. Meaning I don't necessarily trust the solution on Wikipedia about $d(t)$ = $A\cos \omega t$, where $\omega$ = $\sqrt\frac{k}{m}$ after all I have read about pendulums being generally treated with the "small angle approximation".

So, here is what I did:

EDIT: In order to facilitate any differentiation tricks I ensure monotony of all functions by restricting this to a single drop of the spring from A to 0.

(i) $F(d)$ = $-kd$ (Hooke's law)

(ii) $K(d)$=$\int-kx dx$ from $x=d$ to $x=A$. This should be equal to $\frac{k}{2}(A^2-d^2)$.

(iii) $K(v)=\frac{1}{2}mv^2$

(iv) I equate $(ii)$ and $(iii)$ and solve for $d$: $d(v)$ = $\sqrt{A^2-\frac{mv^2}{k}}$

(v) I then differentiate this by $v$: $t$ = $d'(v)$ =$\frac{mv}{k\sqrt{A^2-\frac{mv^2}{k}}}$ (am I crazy?)

(vi) and solve for $v$ to obtain $v(t)$: $v(t)$ =$\frac{Akt}{\sqrt{m(m+kt^2)}}$

(vii) I then insert the expression for $v(t)$ in $(iv)$ to obtain $d(t)$: $d(t)$ = $A\sqrt{\frac{m}{m+kt^2}}$

I'm done, but my result disagrees with Wikipedia. It's quite possible that I have made other mistakes in the process, but judging from the answer that I have now approved, it seems I was much too optimistic anyway, doing t=d'(v). t=d'(v) means "Time is the rate of change of displacement with respect to velocity". Are there any situations at all where this makes sense?

EDIT: In the situastion where a particle starts at displacement $d_0$ and has initial velocity $0$ and is worked upon by a single constant force $f=ma$, then $d(t)=d_0+\frac{1}{2}at^2$. $v(t)=d'(t)=at$, so $t=v(t)/a$, which can be written $t(v)=v/a$, since $v(t)$ is injective. Then $d(v)=d_0+\frac{1}{2a}v^2$ and $d'(v)=v/a=t$. So here it works.

EDIT: This question was inspired by a much easier homework assignment, for which calculating v(t) and d(t) was not even necessary. Having solved the assignment, I was still interested in the theoretical questions described here. In light of all this, I have removed the homework label.

$\endgroup$
12
  • 1
    $\begingroup$ Welcome to Physics.SE! Why would you expect time to be equal to the derivative of displacement with respect to velocity? $\endgroup$ Sep 16, 2015 at 13:40
  • 1
    $\begingroup$ Well... Just as, in the same way as v=d/t for constant values, t=d/v. Then, in some cases v and d can be written as functions of time, and then v is the time derivative of d: v=dd/dt. So when d is written as a function of v, why couldn't t be computed as dd/dv for some problems? $\endgroup$ Sep 16, 2015 at 13:54
  • $\begingroup$ No, t is not in equal to dd/dv. If it did, then dv/dd would equal 1/t and thus we would get dv/ dd * dd/dt = 1/t * dd/dt, which means that dv/dt = v/t. In other words, it would mean that a = v/t. Clearly that's not a statement that's true in general. It would only be true if acceleration was constant and the object's speed at time t=0 was 0. $\endgroup$ Sep 16, 2015 at 14:06
  • $\begingroup$ @Anonymmous The small angle approximation is only used to establish Hook's law. If you use $F= -k x$, you have all ready made the approximation and should not worry about it any more. However, if your force is given by $F=-kx$ (approximation or not), the result from wikipedia is exact. $\endgroup$ Sep 16, 2015 at 14:10
  • $\begingroup$ I can't seem to follow your thoughts, mr Srinivasan. How is this different than the case where you switch v and t, and conclude that dt/dv=1/a? $\endgroup$ Sep 16, 2015 at 14:15

2 Answers 2

0
$\begingroup$

(v) I then differentiate this by $v$: $t$ = $d'(v)$ =$\frac{mv}{k\sqrt{A^2-\frac{mv^2}{k}}}$ (am I crazy?)

You made a mistake, $\frac{d}{dv} d(v)$ doesn't equal time. For instance, $\frac{d}{dv} d(v)$ can be the same at two different times.

Consider the simplest case, a particle is at rest. Then $v=0$ and $d(t)=const $ so the derivative either doesn't exist (you can't change $v$) or else it is zero ($d$ doesn't change) and in neither case is it equal to time.

Consider the next simplest case, $d(t)=vt.$ Now there is no function $d(v)$ because the same $v$ is associated with every possible $d.$

Consider the next simplest case, $d(t)=a+bt+ct^2/2.$ Which has $v(t)=b+ct,$ so $t=(v-b)/c$ which means there is a unique time for each velocity so there is a displacement $d(v)=a+b(v-b)/c+(v-b)^2/2c.$ Differentiating that with respect to velocity gives

$$\frac{d}{dv} d(v)=(b/c)+(v-b)/c=b/c+t\neq t.$$

If it failed for the three simplest cases, there is no real reason to think it works in general, or that it works in your case.

There is a technique called a Legendre Transform that is similar, but you usually don't get to that until later (like in an advanced mechanics course for physics majors) and you generally do it to get a new variable, not to solve for an old one like time. There is no reason to expect that $\frac{d}{dv} d(v)$ takes all values and time should take all values. Time is a parameter you adjust it and get a possibly different configuration and possibly different velocities and momentums and such. Velocity is itself something you get for every time, not every velocity appears even though every time does.

Finally, as a warning, you can't just look at every formula with times and distances and velocities and accelerations and expect it to work just because the units work. Some of those relationships are definitions and some of the concepts look really similar (e.g. average velocity and instantaneous velocity) and some of them are special cases (someone might write $t$ instead of $\Delta t$ because they already choose the zero of time to be one of the two times). Applying a special case in another situation is as wrong as calling everyone by your instructor's name.

So learn the definitions, use good math, and don't do something you can't do and expect it to work.

$\endgroup$
2
  • $\begingroup$ I agree that it should not fail for the three simplest cases, especially not for the third one, where v can take any positive value and the functions v(t) and d(t) are monotonous. I tried with d(t)=sin(t), which returned d'(v)=-cot(t)!=t. Yet another success... I still don't understand it 100 %, though. I will check out the Legendre transform some time. And thanks for the warnings. :-) $\endgroup$ Sep 17, 2015 at 0:32
  • $\begingroup$ By the way I had the plus-minus in front of the square root expression in the original post. I noted it as +-, and then someone removed it while formatting the equations for me... Thanks anyway to whoever fixed the formatting. $\endgroup$ Sep 17, 2015 at 0:43
0
$\begingroup$

Let me use something else instead of $d$ for displacement, say $y$, so we don't get confused with differentials. As it has already been pointed out, you made a mistake at $t=\frac{dy}{dv}$ which is not true in general. I will try to add something to that discussion by saying what it actually is. You motivated your guess by the example of constant acceleration $a$ where you received $t=\frac{v}{a}=\frac{dy}{dv}$. Note that $t=\frac{v}{a}$ is usually only true in this case.

Now let us assume that the function $v(t)$ is invertible to get $t(v)$ (which we don't know). Then we would be able to plug that into our displacement $y(t)$ to get a function for the displacement depending only on $v$, namely $y(t(v))$. Taking the derivative and using the chain rule we get: $$\frac{dy(t(v))}{dv}=\left.\frac{dy(t)}{dt}\right\vert_{t=t(v)}\frac{dt(v)}{dv} $$ Now considering that the derivative of the inverse function is the inverse of the derivative we know that $\frac{dt(v)}{dv}=\frac{1}{a}$ where $a=\frac{dv(t)}{dt}$ is of course the acceleration. Also we know that $v=\frac{dy(t)}{dt}$, so we get in general $\frac{dy(t(v))}{dv}=\frac{v}{a}$. We see that this is not well defined when $a=0$ but that was clear from the beginning because at that point $v(t)$ is not guaranteed to be invertible. Also it is generally not equal to $t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.