Question

Hi all, I have one from a sample exam paper I don't know how to approach. I feel stupid for not knowing but I have tried a few things but am getting nowhere. Here it is: "The mass (m) hanging on an elastic band moves according to the equation [m*(d2s/dt2) = - mg/p(s-L)],
where (s) is the displacement of the mass, (L) is the natural length of the band and (p) is the constant parameter (called reference elongation). Find s as a function of time subject to the initial conditions:

s = s(subnaught), ds/dt = 0 when t = 0.

Any help would be vastly appreciated.

Regards, Rick

Answer 1

Is the (s-L) in the denominator?
It's a little hard to read it properly with how it's written ^^

Answer 2

I have a ss of 60 can I help someone with 99 ss

Answer 3

\[\Large m\cdot s''\quad=\quad -\frac{mg}{p}(s-L)\]
Is this what it's suppose to look like Rick?
Sorry for my ignorance, I'm not very familiar with physics and all that fancy stuff.

Answer 4

Hi Zepdrix the equation you did is exactly the one! The (s-L) is not in the denominator.

Answer 5

Hi "thissucks" I appreciate help from anyone.

Answer 6

Distributing the -mg/p to the terms in the brackets gives us,
\[\Large m s''\quad=\quad -\frac{mg}{p}s+\frac{mgL}{p}\]Adding (mg/p)s to each side gives us,\[\Large m s''+\frac{mg}{p}s\quad=\quad \frac{mgL}{p}\]

Answer 7

Then I guess we would want to put it into standard form (taking the coefficient off of the leading term) so we can ... hmm that would eliminate all of the m's from the equation though... hmm

Answer 8

Either that's telling us that the displacement does not depend on the mass of the object, or I made a mistake somewhere lol :)

Ok let's just continue, I feel like maybe we're on the right track .

Answer 9

This is why I find the question hard as I am not exactly sure what it is asking for, that and I am still an amateur at DE's. It says to find s as a function of time under the initial conditions. if that helps?

Answer 10

The initial conditions will help us `AFTER` we find our general solution.
So let's try to achieve that first.

Dividing through by m gives us,\[\Large s''+\frac{g}{p}s\quad=\quad \frac{gL}{p}\]We'll have to find the homogeneous solution and then the particular after that.

Answer 11

So if we look at the homogeneous case:\[\Large s''+\frac{g}{p}s\quad=\quad 0\]

Answer 12

Are you familiar with solving DE's like this one?
Or at least familiar with what the characteristic equation will look like? :)

In this case our characteristic equation will be like this:\[\Large r^2+\frac{g}{p}=0\]

Answer 13

Ah excellent! I kind of now what is going on know, is this a second order non homogenous de?

Answer 14

I am somewhat familiar. Very new to them.

Answer 15

Ah ok good :) familiar ground!

Answer 16

Yes, it's a Second-Order Linear Non-Homogeneous DE

Answer 17

I know we need to find the overall general solution and that is achieved by setting something to zero and ignoring the rhs, and then finding the particular solution and adding? I am pretty bad at them currently. Which I guess is why I didnt even realise the question was one.

Answer 18

Well the problem involves a ton of constants, very easy to get lost in it.

Answer 19

Is this something you are able to do? If not I completely understand and am already in your debt for helping thus far.

Answer 20

Ya I think we'll be able to get through this one without too much trouble, we'll see how it goes.

Answer 21

So from the characteristic equation:\[\Large r^2+\frac{g}{p}=0\]
Solving for r gives us:\[\Large r\quad=\quad \pm\sqrt{-\frac{g}{p}}\]

Answer 22

Do you recall which type of solution will give us complex roots in the characteristic equation?

Exponentials?
Sines/Cosines?
Logs?

Anything sound familiar? :)

Answer 23

would it be sine/cosines?

Answer 24

I know with exponentials we have to sub the r values into each, well i think so anyway.

Answer 25

I am going to start writing this down so I can refer back to it as this will be a great example, as the couple I did were not as difficult as this.

Answer 26

yes good good.
I guess it technically would be exponentials + sines/cosines.
But the exponential portion comes from the `real` part of our r's.

And you can see that our r's contain no real, just imaginary parts.
So our solution to the homogeneous will contain only sines/cosines.

Answer 27

Ah I see, this is the part were we are trying to guess what will fit our de the best? Am I on the right track?

Answer 28

When:
\[\Large r=\alpha \pm \beta i\]
Our solution is of the form:\[\Large y_h\quad=\quad c_1 e^{\alpha}\cos \beta+c_2 e^{\alpha}\sin \beta\]
^ Just something to remember.

Answer 29

Fit our de the best? +_+ Hmm

Answer 30

Don't feel like you have to hurry and write all this down now.
The thread will be saved in your profile under the `Questions Asked` section.

So you'll always be able to refer back to it.

Answer 31

Oh excellent, I did not know that.

Answer 32

Yeah fit the de, our educated guess, well that's how someone described it to me. Wasn't a very great explanation clearly haha

Answer 33

lol XD

Answer 34

The roots of our characteristic equation gave us:
\[\Large r\quad=\quad 0\pm\sqrt{\frac{g}{p}}i\]

So our homogeneous solution will be of the form:\[\Large y_h\quad=\quad c_1 \cos \left(\sqrt{\frac{g}{p}}\;t\right)+c_2 \sin\left(\sqrt{\frac{g}{p}}\;t\right)\]

Answer 35

And I'm trying to remember ... ummm.. I think we have to find the particular solution BEFORE we can use our initial data. Otherwise we'll run into trouble.

Answer 36

That makes sense as we will then have all the required forms and then we can just sub in the IC?

Answer 37

Yes, and hopefully the IC will allow us to solve for the missing constants, c_1 and c_2, giving us a `specific` solution instead of the general form.

Answer 38

Awesome

Answer 39

\[\Large s''+\frac{g}{p}s\quad=\quad \frac{gL}{p}\]

Our particular solution will be of the same form that the `right side` of the equation has.
Maybe this is what you were referring to by "guessing what will fit our DE best".

Answer 40

yes I think that is! I was shown that we need to match the constants up and then solve it, but I don't remember much else.

Answer 41

Since our `right hand side` is simply constant, the form of the particular solution will also be constant.\[\Large y_p\quad=\quad C\]\[\Large y_p'\quad=\quad 0\]\[\Large y_p''\quad=\quad 0\]
So I set the y_p (particular solution) as an arbitrary constant and then took it's derivative a couple times.

We want to use this list of information and plug it back into the original equation to solve for our constant C.

Answer 42

Oh I'm calling these y's... my bad. I hope that's not too confusing.

Answer 43

I tend to mix up the letters sometimes lol.

Answer 44

no thats all good, is that convention as that is what I have seen? :)

Answer 45

I'm just so used to seeing these types of equations as y''+y'+... instead of involving s.
This is kind of a neat problem though :x cool physics stuff.

So here is our homogeneous part of the solution:
\[\Large s_h\quad=\quad c_1 \cos \left(\sqrt{\frac{g}{p}}\;t\right)+c_2 \sin\left(\sqrt{\frac{g}{p}}\;t\right)\]

I just wanted to fix the y a sec there.

Answer 46

I am glad you are enjoying it :D I am enjoying learning!

Answer 47

This will help me heaps in a couple of weeks when I have my first exam

Answer 48

\[\Large s_p\quad=\quad C\]\[\Large s_p''\quad=\quad 0\]
Plugging these into our original equation gives us:\[\Large s_p''+\frac{g}{p}s_p\quad=\quad \frac{gL}{p}\qquad\quad\to\quad\qquad 0+\frac{g}{p}C\quad=\quad \frac{gL}{p}\]

Answer 49

So what is our particular solution?? :)
It should be jumping out at you! hehe

Answer 50

Is it L? :/ I feel like an idiot. Hold on give me a sec

Answer 51

C = L?

Answer 52

Yes, good good! C=L
And s_p=C is our particular solution.

Answer 53

No that isn't it

Answer 54

oh sweet!

Answer 55

\[\Large s_h\quad=\quad c_1 \cos \left(\sqrt{\frac{g}{p}}\;t\right)+c_2 \sin\left(\sqrt{\frac{g}{p}}\;t\right)\]\[\Large s_h\quad=\quad L\]

And as you mentioned before, our general solution will be the sum of these, yes?\[\Large s\quad=\quad s_h+s_p\]

Answer 56

\[\Large s_p\quad=\quad L\]Grr typo..

Answer 57

Yeah oh awesome, so the general solution will be :

Answer 58

c1cos(gp−−−√t)+c2sin(gp−−−√t) + L

Answer 59

Wait that didn't come out right haha

Answer 60

ah poor guy XD

Answer 61

haha so my answer is then the above which is S(t) = s_p +s_h

Answer 62

brb need some chocolate milk from the kitch:x

Answer 63

That is awesome, that is a very long drawn out problem. So in a physics sense it is telling me then that the displacement as a function of time is governed by the length, obviously, and the constant parameter.

Answer 64

haha okay, get me a glass while you're there ;D

Answer 65

Ok good so the sum of our parts gives us:\[\Large s(t)\quad=\quad c_1 \cos \left(\sqrt{\frac{g}{p}}\;t\right)+c_2 \sin\left(\sqrt{\frac{g}{p}}\;t\right)+L\]
Something like that, right?

Answer 66

And our initial conditions are:\[\Large s(0)\quad=\quad s_o\]\[\Large s'(0)\quad=\quad0\]

Answer 67

Yep that is them, so we don't need to do anything else with it do we? As with s(0) = s_0 would just equal L?

Answer 68

Hmm let's see.. I think it gives us a little bit more than that.

Answer 69

oh wait one sec

Answer 70

if t = 0 then it would be ccos(0) + csin(0) + L

Answer 71

Hmm cos(0)=?

Answer 72

Wait... that was stupid sorry

Answer 73

C_1 + C_2 + L as cos (0) is 1? I am most probably wrong

Answer 74

My thinking was that due to t = 0 that would cancel out the fraction inside the cos (*)

Answer 75

Ya it kills the fraction inside of cosine and sine, making 0's inside those spots.
cos(0)=1, sin(0)=0.

So it looks like s(0) is telling us something about c_1, right?\[\Large s_o\quad=\quad c_1 \cos(0)+\cancel{c_2\sin(0)}+L\]

Answer 76

oh damn sorry I completely forgot the sin!

Answer 77

so if we rearrange for C_1

Answer 78

it is equal to 1? or s(t) - L

Answer 79

Ya you're on the right track, your notation is a little sloppy though ^^
\[\Large s(0)\quad=\quad s_o\quad = \quad c_1+L\]So solving for c_1 gives us:\[\Large c_1\quad=\quad s_o-L\]

Answer 80

Yeah it is terrible, I haven't really done the whole typing equations out a lot, I am a pen and paper guy haha

Answer 81

Ya getting math help online can be a real pain lol.
I LOVE this equation tool now that I've finally gotten used to it.
Takes some time though.

Answer 82

I will definitely have to keep using it as the help you have given today is better than I got at uni from a guy that gets paid, you should do tutoring if you already are not.

Answer 83

So can we do anything else with it? We now have the constant, there is not much more we could do is there? Or maybe we need to take the derivative? I don't know

Answer 84

Ya I have an interview new week actually :) Trying to find some tutoring work while I'm in school.

This math is all fresh in my mind so it's a good time to try.

Answer 85

Yes.
So let's see if the other initial condition tells us anything about c_2.

Answer 86

Derivative time.

Answer 87

\[\Large s(t)\quad=\quad c_1 \cos \left(\sqrt{\frac{g}{p}}\;t\right)+c_2 \sin\left(\sqrt{\frac{g}{p}}\;t\right)+L\]
\[\Large s'(t)\quad=\quad ?\]

Answer 88

Don't forget your chain rule!
This one won't be too bad.

Answer 89

be one sec :)

Answer 90

I think

Answer 91

Did it go through?

Answer 92

If it's too much of a pain to type out, that's fine. I can do it lol

Answer 93

It didn't :OOO

Answer 94

nah ill type it out again only be a sec

Answer 95

\[\sqrt{g/p}(\cos(t \sqrt{g/p})) - \sin(t \sqrt{g/p})\]

Answer 96

I think thats it.

Answer 97

I know i should know derivatives of cos and sin and the chain rule inside out, but I am a bit rusty.

Answer 98

Ok looks pretty close:\[\Large s'(t)\quad=\quad \color{red}{c_2}\sqrt{g/p}(\cos(t \sqrt{g/p})) -\color{red}{c_1}\color{red}{\sqrt{g/p}}\sin(t \sqrt{g/p})\]