Question

A body of mass m kg attached to a spring moves with friction. The motion is described by the second Newton's law: m(d^2y/dt^2) + a(dy/dt) + ky = 0 where y is the body displacement in m, t is the time in s, a > 0 is the friction coeffcient in kg/s and k is the spring constant in kg/s2. Assuming m = 1 kg and k = 4 kg/s2, find:

Answer 1

a) What is the range of values of a for which the body moves (i) with oscillations, (ii) without oscillations? b) Find the general solution for any a < 4. (Your solution should be a formula depending on the parameter a.) Proof that it follows from the solution obtained that the body slows down to a virtually rest state at large time (i.e. when t --> infinity)? c) Find the particular solution for a = 4 subject to the initial conditions y(0) = 0, dy/dt = 1 m/s at t = 0. Plot this solution and determine the largest displacement of the mass using calculus.

Answer 2

This is a standard non-homogeneus second order ODEs with constant coefficients, is it not?

Answer 3

homogeneus second order ODE with constant coefficients

Answer 4

now you have to solve characteristic polynomial mt^2+at+ky=0

Answer 5

to get y by itself on the left?

Answer 6

sorry
mt^2+at+k=0

Answer 7

now you will have solutions for t..
you know how the solution for the differential equation look if you have one t
two t or minus under the square root ?

Answer 8

right which in our case we have one t

Answer 9

we better write the equation as mw^2+aw+k=0
because t might confuse us with the time .. we really find from this equation the angular frequency

Answer 10

in (a) they want you to find the range of values for a ..

Answer 11

oscillations will occur only when there is negative sign under the square root

Answer 12

yep ok as oscillations require +'s and -'s. So as we have no -'s for this question?

Answer 13

what do you mean ?

Answer 14

mw^2+aw+k=0

for oscillations :
a^2-4mk < 0

Answer 15

i have to go now but ill come back later

Answer 16

ok thanks cause Im stuck with this one.

Answer 17

im back.. any progress ?

Answer 18

no sorry Ive been working on other problems. I had a go but am really stuck on this one.

Answer 19

thanks for coming back!

Answer 20

so we said for oscillations we need to have

a^2-4mk < 0
a^2 < 16
-4 < a < 4

Answer 21

since a > 0 we have 0< a < 4
agree so far ?

Answer 22

for a => 4 we have no oscillations

Answer 23

Yes following so far

Answer 24

so done with a) now go to b)

the general solution (using the fact that we have negative under the square root) is given by:

\[y=e^{\lambda \times t}(c1Sin(\omega \times t) + c2Cos(\omega \times t))\]

Answer 25

where

\[\lambda \pm i \mu = \frac{ -a \pm \sqrt{a^2-4mk} }{ 2m }\]

Answer 26

now since we know a<4 -> a^2<4mk
we can re write the last thing i wrote as :

\[\lambda \pm i \mu = \frac{ -a }{ 2m } \pm i\frac{ \sqrt{4mk - a^2} }{ 2m }\]

Answer 27

plug in values gives us :

\[\lambda \pm i \mu = \frac{ -a }{ 2 } \pm i\frac{ \sqrt{16-a^2} }{ 2 }\]

Answer 28

agree with what i did so far ?

Answer 29

where did the equation y=eλ×t(c1Sin(ω×t)+c2Cos(ω×t)) come from?

Answer 30

this is the general solution if you have minus under the square root..
you have 3 cases.. for each case the general solution looks different

Answer 31

You mean in the equation \[a^2-4mk\]

Answer 32

no for the whole differential equation ..
the cases are determined by a^2 -4mk .. it might be 0, larger than 0 or smaller than 0

Answer 33

http://www.sosmath.com/diffeq/second/constantcof/constantcof.html

Answer 34

Oh a quadratic equation. following you now.

Answer 35

the discriminant of the quadratic equation tells you how the solution of the differential equation will look like

Answer 36

so you can use the general solution for a<4 that i gave you and plug in the
lambda and mu that i wrote and there you have the general for a<4.

Answer 37

ok I remember doing similar problems as per the reference. You may continue Im with you now.

Answer 38

so you can use the general solution for a<4 that i gave you and plug in the
lambda and mu that i wrote and there you have the general for a<4.

Answer 39

Im not following what you mean here

Answer 40

i gave you a general solution
y=eλ×t(c1Sin(ω×t)+c2Cos(ω×t))
and then i showed you what are λ and ω (i showed you what are λ and mu instead of mu i should have written ω its just a mistake that i wrote mu)

Answer 41

look at the last thing i wrote before you asked
"where did the equation y=eλ×t(c1Sin(ω×t)+c2Cos(ω×t)) come from?"

Answer 42

just treat mu as omega and plug it into the general solution

Answer 43

whats plugged in for a as we have a>4? any number between 0-4?

Answer 44

we have a <4

Answer 45

sorry typo

Answer 46

and we should leave a as a .. they want an equation with a

Answer 47

so you can choose any a you want

Answer 48

as long as it between 0 and 4

Answer 49

"Your solution should be a formula depending on the parameter a"

Answer 50

So I have to plug in equation for λ into the general solution above?

Answer 51

replace lambda by -a/2
replace omega by sqrt(16-a^2)/2

Answer 52

that ends up being a big equation. What am I solving it for, a?

Answer 53

no.. this is the general solution for a<4..

Answer 54

you dont have to do anything more

Answer 55

if they told you what a is
and some more initial conditions
you could say what is 'y' for every 't' ..

Answer 56

we found an equation that tells the position with respect to time

Answer 57

So this is the final answer for part (b)
\[y=e ^{\frac{ -a }{ ?2}xt}c1Sin(\sqrt{\frac{ 16-a ^{2} }{ 2 }}\times t)+c2Cos(\sqrt{\frac{ 16-a ^{2} }{ 2 }}\times t)\]

Answer 58

you know you can remove the "x" sign its just multiplication
and the 2 is not under the square root

Answer 59

and e^(-at/2) multiplies both c1sin.. and c2cos..

Answer 60

Why is the 2 not under the square root?

Answer 61

look at the solution up there

Answer 62

follows from the solution to quadratic equation

Answer 63

Ah yes I see its just a typo in me writing the equation in this forum. I have written it correctly on paper. And I also missed the brackets after the e^(-at/2) but I have written it down correctly.

Answer 64

And lastly part (c)
when a=4
Auxiliary equation is;
p^2+4p+4=0

Which both roots equal -2, therefore general solution y(t)=(At+B)e^kt where k=-2

Initial conditions y(0)=0, y'(0)=1

Therefore B=1, -2B+A=1 hence A=3

Therefore final solution;
y(t)=(3t+1)e^-2t

Answer 65

i think B should be 0

Answer 66

Hmm why?

Answer 67

y(0) = 0

Answer 68

So my final solution is wrong?

Answer 69

y(0) = 0
means B = 0

y(t) = (At+B)e^kt
y'(t) = Ae^kt + kAte^k(t)
y'(0) = A = 1

y(t)=te^-2t

Answer 70

Wow thanks so much Coolsector. Your patience and time has been much appreciated. I will go away and practice what you have showed me.

Answer 71

yw :) good luck with that
you should remember there are 3 cases .. for each one the solution looks different
it all follows with "assuming" a solution of y = Ae^(iwt)

Answer 72

Oh and the last bit of (c) is to sketch solution and determine largest displacement of the mass. We have done the last part havent we? I just have to do the sketch?

Answer 73

where in the general case A can be complex as well

Answer 74

sorry sorry misrake

Answer 75

and the plot is just an upside down parabola?

Answer 76

you have to plot this : y(t)=te^-2t
in order to find larger displacement you should take the derivative
y'(t) = e^-2t -2te^-2t
equate to 0
(1-2t)e^(-2t) = 0
t = 0.5
this is the maximum for sure since this function rises at the beginning then decreasing to 0 as t goes to infinity
now plug y(0.5) to get maximum displacement

Answer 77

no! this is not a parabola!!!
t*e^(-2t)

Answer 78

http://www.wolframalpha.com/input/?i=te%5E%28-2t%29+from+0+to+5

Answer 79

wow great reference showing the plot as well! I will bookmark that!

Answer 80

wolframalpha is a really good tool :)

Answer 81

you can as well plug in there te^(-2t) and it gives you information about the function
like maximum, limits etc

Answer 82

anyway i have to go now .. see you.. good luck

Answer 83

Yes many thanks you have been a lifesaver. Have a good day!