Newton's second law for a falling object

**AAKhan07** · April 16th 2009, 09:05 AM

Hi, the two forces acting on an object are its weight and air resistance. In my differential equations coursework I begin with 'air resistance = -kv' where v is the velocity and k is a constant. This is easy enough to solve.
$ m\frac{d^2s}{dt^2} = mg - kv $

$ m\frac{d^2s}{dt^2} + kv = mg $

$ m\frac{d^2s}{dt^2} + k\frac{ds}{dt} = mg $

Findgin auxiliary equation
$ m\lambda^2 +k\lambda = 0 $

$ \lambda(m\lambda + k) = 0 $

$ \lambda = 0 or \lambda = \frac{-k}{m} $

$ \lambda = \frac{-k}{m} $

$ s=A+Be^\frac{-kt}{m} $

Finding particular integral
$ s=at $

$ \frac{ds}{dt} =a $

$ \frac{d^2s}{dt^2} =0 $

$ m(0) + k(a) = mg $

$ a = \frac{mg}{k} $

$ s= \frac{mg}{k}t $

General Solution

$ s = A+Be^\frac{-kt}{m} + \frac{mg}{k}t $

$ s=0 $
when
$ t=0 $

$ s = A+Be^\frac{-k(0)}{m} + \frac{mg}{k}(0) $

$ 0 = A+B $

$ A = -B $

Therefore

$ s = -B+Be^\frac{-kt}{m} + \frac{mg}{k}t $

$ \frac{ds}{dt}=0 $
when
$ t=0 $

$ \frac{ds}{dt}=\frac{-kB}{m}e^\frac{-kt}{m} + \frac{mg}{k} $

$ 0=\frac{-kB}{m} + \frac{mg}{k} $

$ B=\frac{m^2g}{k^2} $

So the equation is

$ s = -\frac{m^2g}{k^2}+\frac{m^2g}{k^2}e^\frac{-kt}{m} + \frac{mg}{k}t $

I would appreciate if you could first check the correctness of the above working.
My main question is: how do you find the differential equation when the air resistance = -kv^2
This time the air resistance is dependent on the square of the speed rather than the speed. Newton's second law becomes:

$ m\frac{d^2s}{dt^2} = mg - kv^2 $

I am looking to solve this for s as I did for the other equations.

**Danny** · April 16th 2009, 11:37 AM

Originally Posted by AAKhan07

Hi, the two forces acting on an object are its weight and air resistance. In my differential equations coursework I begin with 'air resistance = -kv' where v is the velocity and k is a constant. This is easy enough to solve.
$ m\frac{d^2s}{dt^2} = mg - kv $

$ m\frac{d^2s}{dt^2} + kv = mg $

$ m\frac{d^2s}{dt^2} + k\frac{ds}{dt} = mg $

Findgin auxiliary equation
$ m\lambda^2 +k\lambda = 0 $

$ \lambda(m\lambda + k) = 0 $

$ \lambda = 0 or \lambda = \frac{-k}{m} $

$ \lambda = \frac{-k}{m} $

$ s=A+Be^\frac{-kt}{m} $

Finding particular integral
$ s=at $

$ \frac{ds}{dt} =a $

$ \frac{d^2s}{dt^2} =0 $

$ m(0) + k(a) = mg $

$ a = \frac{mg}{k} $

$ s= \frac{mg}{k}t $

General Solution

$ s = A+Be^\frac{-kt}{m} + \frac{mg}{k}t $

$ s=0 $
when
$ t=0 $

$ s = A+Be^\frac{-k(0)}{m} + \frac{mg}{k}(0) $

$ 0 = A+B $

$ A = -B $

Therefore

$ s = -B+Be^\frac{-kt}{m} + \frac{mg}{k}t $

$ \frac{ds}{dt}=0 $
when
$ t=0 $

$ \frac{ds}{dt}=\frac{-kB}{m}e^\frac{-kt}{m} + \frac{mg}{k} $

$ 0=\frac{-kB}{m} + \frac{mg}{k} $

$ B=\frac{m^2g}{k^2} $

So the equation is

$ s = -\frac{m^2g}{k^2}+\frac{m^2g}{k^2}e^\frac{-kt}{m} + \frac{mg}{k}t $

I would appreciate if you could first check the correctness of the above working.
My main question is: how do you find the differential equation when the air resistance = -kv^2
This time the air resistance is dependent on the square of the speed rather than the speed. Newton's second law becomes:

$ m\frac{d^2s}{dt^2} = mg - kv^2 $

I am looking to solve this for s as I did for the other equations.

It's really easier if you let $\frac{ds}{dt} = v$ in your first problem. It becomes first order and separable. This also works for your second problem.

**running-gag** · April 16th 2009, 11:48 AM

Hi

Your work is OK even if it can be done maybe a little bit more easily by solving the differential equation with v instead of s

$m\frac{d^2s}{dt^2} = mg - kv$

$m\frac{d^2s}{dt^2} + kv = mg$

$m\frac{dv}{dt} + kv = mg$

The general solution is $v_g = A\:e^{-\frac{k}{m}\:t}$

A particular solution is $v_p = \frac{mg}{k}$

The solution is therefore $v = v_g + v_p = A\:e^{-\frac{k}{m}\:t} + \frac{mg}{k}$

$v=0$ when $t=0$ gives $v = -\frac{mg}{k}\:e^{-\frac{k}{m}\:t} + \frac{mg}{k}$

Integrating gives $s = \frac{m^2g}{k^2}\:e^{-\frac{k}{m}\:t} + \frac{mg}{k}\:t + B$

$s=0$ when $t=0$ gives $s = -\frac{m^2g}{k^2}+\frac{m^2g}{k^2}\:e^{-\frac{k}{m}\:t} + \frac{mg}{k}\:t$

**AAKhan07** · April 16th 2009, 12:08 PM

Thanks, I didn't realize I was overcomplicating it, although I'm still not sure about how to solve the second equation. It should be exactly the same except with a different auxiliary equation (or general equation, as running-gag called it). I'm not sure what the auxiliary equation is for the second case, any of you know?

**running-gag** · April 16th 2009, 01:17 PM

Yes

$m\frac{d^2s}{dt^2} = mg - kv^2$

$m\frac{dv}{dt} = mg - kv^2$

$\frac{m\:dv}{mg - kv^2} = dt$

Let $v = \sqrt{\frac{mg}{k}} w$

$\sqrt{\frac{m}{kg}}\:\frac{dw}{1 - w^2} = dt$

Integrating gives $\sqrt{\frac{m}{kg}}\:argtanh(w) = t$

$w = tanh\left(\sqrt{\frac{kg}{m}}t\right)$

$v = \sqrt{\frac{mg}{k}}\: w = \sqrt{\frac{mg}{k}} \:tanh\left(\sqrt{\frac{kg}{m}}t\right)$

**AAKhan07** · April 16th 2009, 01:46 PM

thanks, that's a good solution, can you get it with s as the subject instead of v? Im doing the experiment at home so I can't really measure speeds, only distances and times.

**running-gag** · April 17th 2009, 09:13 AM

$\frac{ds}{dt} = \sqrt{\frac{mg}{k}} \:tanh\left(\sqrt{\frac{kg}{m}}\:t\right)$ gives $s = \frac{m}{k} \:\ln\left(cosh\left(\sqrt{\frac{kg}{m}}\:t\right) \right)$

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