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Fall with drag

  • Thread starter luckis11
  • Start date
  • #1
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x’’(t)=g-Fd/m=(9.81metres/sec^2)-(k/m)(x’(t))
where x''(t)=du/dt and x'(t)=u=dx/dt
I enter this at http://www.wolframalpha.com/input/?i=
and I get the solution x(t)= -(c1me^(-kt/m))/k+c2+9.81(m/k)t
But
x(t=0)=0
x’(t=0)=0
x(t=0)=c1+c2
x’(t=0)=c1-c2
=>c1=c2=0
=>x(t)=9.81(m/k)t
and this says that it reached terminal speed right from the start of the fall, and it is not the solution mentioned at H.Young p.126. Where's the mistake?
 
Last edited:

Answers and Replies

  • #2
51
0
First of all, x(t)=9.81(m/k)t is totally wrong, because it states that the velocity is constant, which means that the acceleration is zero.

Here is the solution:

Set y=x', then you have the following 1st order eqn:

y'+(k/m)*y=g => y' / (g-(k/m)y)=1 => y=(m/k)*(g-c1*exp(-k*t/m)), where c1 is constant.

Thus, x'=(m/k)*(g-c1*exp(-k*t/m))=> x=(m/k)*g*t-(m/k)^2*exp(-k*t/m)+c2

I assume that at t=0, x'=0 =>c1=g
and at t=0, x=0 => c2=(m/k)^2

For the terminal velocity, you need to have x''=0 ==> g-(k/m)x'=0, which gives t->oo

What are the initial conditions?
 
  • #3
230
0
Greatfull for your will to help me, but I did not ask for the correct solution. The correct solution is supposed to be H.Young's p.126, which I have it. I asked where is the mistake in the formulations I stated.

I made a mistake at my previous post, I edited and corrected it now, read it again if you want. I missed to mention that the initial conditions are:
x(t=0)=0, x’(t=0)=0.

Is the mistake that the
x(t=0)=c1+c2
x’(t=0)=c1-c2
is wrong? I was told in this forum that this is so. If it is wrong then how am I supposed to calculate c1 and c2 when I am given a 2nd order dif. eq. solution by the calculator which contains c1 and c2?
 
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  • #4
51
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Well, you have made several mistakes:

x(t=0)=0 does not give you c1+c2=0, but -c1*(m/k)+c2=0
also x'(t=0)=c1+9.81(m/k)
 
  • #5
230
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I was told or thought I was told that in any case:
x(t=0)=c1+c2, x’(t=0)=c1-c2
thus since x(t=0)=0, x’(t=0)=0,
it is c1+c2=0, c1-c2=0,
thus c1=c2=0.

So I guess from what you now said, this is wrong. And my question is, what is the correct procedure to find c1 and c2 when I see a 2nd order dif. eq. solution at the calculator of the form:

x(t)=(here there are c1 and c2).
 
  • #6
51
0
Have you taken any course on differential equations before?

Anyway,
you have the following expression:

x(t)= -(c1me^(-kt/m))/k+c2+9.81(m/k)t

as you can see, "t" appears on both sides of the equation.
The first initial condition states that x(t=0)=0. Therefore, in the above equation set t=0. You also know that x(t=0)=0, which means that the left hand side of the eqn is 0. Thus you get

0=-(c1m)/k+c2

In order to apply the 2nd initial condition, you need to differentiate the equation with respect to t. Then, as before you set t=0, and you know that the left hand side of the new equation is 0. Thus,

x'(t)=c1e^(-kt/m)+9.81(m/k)
and x'(t=0)=0 => 0=c1+9.81(m/k)
 
  • #7
230
0
Greatfull.:smile:
 

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