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When analyzing the movement of a weight attached to a spring, many sources set up the force equation using newton’s second law as follows. $$mg-k(L+x)=ma$$ where $L$ is the length that the mass $m$ stretches the string from its relaxed position.

There always has to be a negative sign in front of the spring force because it always moves opposite the sign of the displacement $x$. I guess $mg$ is positive because it’s conventional to designate the downward direction as positive.

But the thing is, I’ve been told that your choice of coordinate system doesn’t matter. So what if we pick the upwards direction as being positive? Well then we get $$-mg-k(L+x)=\pm ma$$ where I’m not sure what the sign on $ma$ is. It turns out it doesn’t matter what it is because this equation is wrong regardless! Solving this differential equation in either case produces a solution that is not of the form $Asin(\omega t+\phi)$.

Do you have to choose a coordinate system that points downward? If so why? If not, what are you supposed to do for a coordinate system where gravity points downward like normal?

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  • $\begingroup$ Changing the constant term from negative to positive was enough to change the form of the DE solution that much? $\endgroup$
    – JMac
    Commented Sep 21, 2019 at 1:23
  • $\begingroup$ I think you mean positive to negative, and yes it was. It’s because the $mg$ and $kL$ terms are supposed to cancel out. $\endgroup$
    – Ryan
    Commented Sep 21, 2019 at 1:42
  • $\begingroup$ All you need to do is to use F=ma and remember to be consistent about which direction is positive. Your 1st equation is correct. The expression on the left side of your 2nd equation for the force is correct. So just use F=ma, which means that you should use the "+" sign in eqn. 2. This equation does then have a sinusoidal solution since the equation basically becomes -kx=ma (or -kx=mx'') after cancelling out the constant terms in the equation. You get the exact same equation with your 1st equation after cancelling out the constant terms. $\endgroup$
    – user93237
    Commented Sep 21, 2019 at 2:51
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    $\begingroup$ @RyanGreyling If the sign of your $g$ changes, what about the sign of your $L$? $\endgroup$
    – JMac
    Commented Sep 21, 2019 at 2:54
  • $\begingroup$ @SamuelWeir No, you wouldn’t get the same solution. The solution in that case is $Asin(\omega t+\phi)+\frac{mg+kL}{k}$ $\endgroup$
    – Ryan
    Commented Sep 21, 2019 at 3:07

1 Answer 1

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m

The equation of motion can't depend on your choice of a coordinate system.

How to get the equation of motion

I) choose arbitrary coordinate (positive y direction, $\dot{y}$ and $\ddot{y}$)

II) the inertia force ($m\,a$) is always opposite to the acceleration $\ddot{y}$

III) because the weight force ,$m\,g$ cause the mass to move toward the y direction, put this force toward the y direction.

IV) the spring force $F_C$ is a "cut" force , so according to Newton 3'rd law , you can put it toward the y direction or opposite to the y direction.

the sum of these forces give you the equation of motion.

Case I

$$\sum_{+y}=0=m\,g-m\,a+F_c=0$$

$\Rightarrow$

$$m\,\,a-F_c=m\,g\tag 1$$

Case II $$\sum_{+y}=0=m\,g-m\,a-F_c=0$$

$\Rightarrow$

$$m\,a+F_c=m\,g\tag 2$$

to get harmonic oscillation the spring force must be in equation (1)

$F_c=k\,(L-x)$

and for equation (2)

$F_c=k\,(x-L)$

conclusion:

the equation of motion (1) and (2) are equal, even if you choose different coordinate system

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  • $\begingroup$ Those equations are wrong though. If you tried solving those differential equations you'd get $Asin(\omega t + \phi)$ plus a constant term. You need to flip the sign of the L in both equations. Also, I tried googling "cut force" but nothing came up. $\endgroup$
    – Ryan
    Commented Sep 24, 2019 at 4:35
  • $\begingroup$ @RyanGreyling this is your equation: $m{\frac {d^{2}}{d{t}^{2}}}x \left( t \right) +k \left( x \left( t \right) -L \right) =mg $ and this is the solution $x(t)=-\cos \left( {\frac {\sqrt {k}t}{\sqrt {m}}} \right) \left( kL+mg \right) \frac{1}{k}+{\frac {kL+mg}{k}} $ initial condition $x(0)=0$ and $D(x)(0)=0$ $\endgroup$
    – Eli
    Commented Sep 24, 2019 at 8:18

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