# Thread: Newton's second law for a falling object

1. ## Newton's second law for a falling object

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

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

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

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

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

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

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

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

Finding particular integral
$\displaystyle s=at$

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

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

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

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

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

General Solution

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

$\displaystyle s=0$
when
$\displaystyle t=0$

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

$\displaystyle 0 = A+B$

$\displaystyle A = -B$

Therefore

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

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

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

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

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

So the equation is

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

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:

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

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

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

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

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

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

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

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

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

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

Finding particular integral
$\displaystyle s=at$

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

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

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

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

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

General Solution

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

$\displaystyle s=0$
when
$\displaystyle t=0$

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

$\displaystyle 0 = A+B$

$\displaystyle A = -B$

Therefore

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

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

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

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

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

So the equation is

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

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:

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

I am looking to solve this for s as I did for the other equations.
It's really easier if you let $\displaystyle \frac{ds}{dt} = v$ in your first problem. It becomes first order and separable. This also works for your second problem.

3. 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

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

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

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

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

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

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

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

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

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

4. 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?

5. Yes

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

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

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

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

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

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

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

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

6. ## Good solution

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.

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