Transforming a differential equation into another differential equation

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May 8th 2011, 07:40 AM
Basheesh

Transforming a differential equation into another differential equation

1. The problem statement

$m\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=F_{0}cos \omega t \\\\LC\frac{d^{2}V_{c}}{dt^{2}}+RC\frac{dV_{c}}{dt }+V_{c}=V_{0}cos\omega t$

These two equations are for two different physical phenomena. The first is for a mass moving along an x axis. The mass is affected by a springforce k, a dampener with constant c and an outer force F(t), here set to $F_0cost(\omega t)$ .
Explanation of the second equation follows. The gist is that they are mathematically the same (I'm translating from danish, let me know if you need the rest).

a) show that both differential equations can be brought to the form:

$\frac{dx^{2}}{dt^{2}}+2\alpha \frac{dx}{dt}+\omega_{0}^{2}x=fcos\omega t$

Where $\alpha \geq 0$ and we assume $\omega > 0$

2. What I've got

I'm not even sure where to start with this.
So I'm thinking that I somehow need to use:
$\omega_{0}^{2} = c^2 + 4mk$

and possible take the integral of the whole thing. But I am pretty lost.
May 8th 2011, 07:44 AM
Matt Westwood

Quote:

Originally Posted by Basheesh

1. The problem statement

$m\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=F_{0}cos \omega t \\\\LC\frac{d^{2}V_{c}}{dt^{2}}+RC\frac{dV_{c}}{dt }+V_{c}=V_{0}cos\omega t$

These two equations are for two different physical phenomena. The first is for a mass moving along an x axis. The mass is affected by a springforce k, a dampener with constant c and an outer force F(t), here set to $F_0cost(\omega t)$ .
Explanation of the second equation follows. The gist is that they are mathematically the same (I'm translating from danish, let me know if you need the rest).

a) show that both differential equations can be brought to the form:

1:
$\frac{dx^{2}}{dt^{2}}+2\alpha \frac{dx}{dt}+\omega_{0}^{2}x=fcos\omega t$

Where $\alpha \geq 0$ and we assume $\omega > 0$

2. What I've got

I'm not even sure where to start with this.
So I'm thinking that I somehow need to use:
$\omega_{0}^{2} = c^2 + 4mk$

and possible take the integral of the whole thing. But I am pretty lost.

Looks like in 1. you need to divide through by m, and this makes the 2 \alpha coefficient c/m, and \omega_0^2 equal to k / m, and f is F_0 / m.

Then in 2. you divide through by LC and do similar.
May 8th 2011, 07:46 AM
Basheesh

Quote:

Originally Posted by Matt Westwood

Looks like in 1. you need to divide through by m, and this makes the 2 \alpha coefficient c/m, and \omega_0^2 equal to k / m, and f is F_0 / m.

Then in 2. you divide through by LC and do similar.

But what about the $\frac{d^2x}{dt^2}$ turning into $\frac{dx^2}{dt^2}$ then?
May 8th 2011, 07:58 AM
Basheesh

Quote:

Originally Posted by Basheesh

But what about the $\frac{d^2x}{dt^2}$ turning into $\frac{dx^2}{dt^2}$ then?

I just talked to a friend about this, and we concluded that it must be a typo, in which case your solution is sufficient. Does anyone agree/disagree with this conclusion? Thanks!
May 8th 2011, 12:27 PM
Matt Westwood

Quote:

Originally Posted by Basheesh

I just talked to a friend about this, and we concluded that it must be a typo, in which case your solution is sufficient. Does anyone agree/disagree with this conclusion? Thanks!

Correct. In this context, $\frac {dx^2}{dt^2} = \left({\frac {dx}{dt}}\right)^2$ is incorrect if not meaningless. Sorry, I didn't comment on it, thought it was a transcription error.