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Math Help - Newton's second law for a falling object

  1. #1
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    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.
    <br />
m\frac{d^2s}{dt^2} = mg - kv<br />

    <br />
m\frac{d^2s}{dt^2} + kv = mg<br />

    <br />
m\frac{d^2s}{dt^2} + k\frac{ds}{dt} = mg<br />

    Findgin auxiliary equation
    <br />
m\lambda^2 +k\lambda = 0<br />

    <br />
\lambda(m\lambda + k) = 0<br />

    <br />
\lambda = 0 or \lambda = \frac{-k}{m}<br />

    <br />
\lambda = \frac{-k}{m}<br />

    <br />
s=A+Be^\frac{-kt}{m}<br />

    Finding particular integral
    <br />
s=at<br />

    <br />
\frac{ds}{dt} =a<br />

    <br />
\frac{d^2s}{dt^2} =0<br />

    <br />
m(0) + k(a) = mg<br />

    <br />
 a = \frac{mg}{k}<br />

    <br />
 s= \frac{mg}{k}t<br />

    General Solution


    <br />
s = A+Be^\frac{-kt}{m} + \frac{mg}{k}t<br />

    <br />
s=0<br />
    when
    <br />
t=0<br />

    <br />
s = A+Be^\frac{-k(0)}{m} + \frac{mg}{k}(0)<br />

    <br />
0 = A+B<br />

    <br />
 A = -B<br />

    Therefore

    <br />
s = -B+Be^\frac{-kt}{m} + \frac{mg}{k}t<br />

    <br />
 \frac{ds}{dt}=0<br />
    when
    <br />
 t=0<br />

    <br />
  \frac{ds}{dt}=\frac{-kB}{m}e^\frac{-kt}{m} + \frac{mg}{k}<br />

    <br />
  0=\frac{-kB}{m} + \frac{mg}{k}<br />

    <br />
   B=\frac{m^2g}{k^2}<br />

    So the equation is

    <br />
s = -\frac{m^2g}{k^2}+\frac{m^2g}{k^2}e^\frac{-kt}{m} + \frac{mg}{k}t<br />

    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:

    <br />
m\frac{d^2s}{dt^2} = mg - kv^2<br />

    I am looking to solve this for s as I did for the other equations.
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  2. #2
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    Quote Originally Posted by AAKhan07 View Post
    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.
    <br />
m\frac{d^2s}{dt^2} = mg - kv<br />

    <br />
m\frac{d^2s}{dt^2} + kv = mg<br />

    <br />
m\frac{d^2s}{dt^2} + k\frac{ds}{dt} = mg<br />

    Findgin auxiliary equation
    <br />
m\lambda^2 +k\lambda = 0<br />

    <br />
\lambda(m\lambda + k) = 0<br />

    <br />
\lambda = 0 or \lambda = \frac{-k}{m}<br />

    <br />
\lambda = \frac{-k}{m}<br />

    <br />
s=A+Be^\frac{-kt}{m}<br />

    Finding particular integral
    <br />
s=at<br />

    <br />
\frac{ds}{dt} =a<br />

    <br />
\frac{d^2s}{dt^2} =0<br />

    <br />
m(0) + k(a) = mg<br />

    <br />
a = \frac{mg}{k}<br />

    <br />
s= \frac{mg}{k}t<br />

    General Solution


    <br />
s = A+Be^\frac{-kt}{m} + \frac{mg}{k}t<br />

    <br />
s=0<br />
    when
    <br />
t=0<br />

    <br />
s = A+Be^\frac{-k(0)}{m} + \frac{mg}{k}(0)<br />

    <br />
0 = A+B<br />

    <br />
A = -B<br />

    Therefore

    <br />
s = -B+Be^\frac{-kt}{m} + \frac{mg}{k}t<br />

    <br />
\frac{ds}{dt}=0<br />
    when
    <br />
t=0<br />

    <br />
\frac{ds}{dt}=\frac{-kB}{m}e^\frac{-kt}{m} + \frac{mg}{k}<br />

    <br />
0=\frac{-kB}{m} + \frac{mg}{k}<br />

    <br />
B=\frac{m^2g}{k^2}<br />

    So the equation is

    <br />
s = -\frac{m^2g}{k^2}+\frac{m^2g}{k^2}e^\frac{-kt}{m} + \frac{mg}{k}t<br />

    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:

    <br />
m\frac{d^2s}{dt^2} = mg - kv^2<br />

    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.
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  3. #3
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    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
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  4. #4
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    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?
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  5. #5
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    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)
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  6. #6
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    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.
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  7. #7
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    \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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