Spring-Mass System: Treating 2 Identical Springs

[colonel]

Hi, in a simple spring-mass system consisting of two identical springs, how would you treat the springs? Would Hookes equation be F = kx + kx?

 

[Tide]

Are they in series or parallel?

 

[recon]

You might want to take a look here: http://online.cctt.org/physicslab/content/phyapb/lessonnotes/springs/lessonsprings.asp

 

[HallsofIvy]

Tides point is that if you attach two identical springs end to end, effectively you still have just one spring (of twice the length) still with spring constant k.

If you have two identical springs side by side (and both attached to the mass) then the act identically and then you can add them.

 

[Tide]

Tides point is that if you attach two identical springs end to end, effectively you still have just one spring (of twice the length) still with spring constant k.

If you have two identical springs side by side (and both attached to the mass) then the act identically and then you can add them.

Exactly - except - the spring constant of a spring varies inversely with its length! This means that placing two identical springs in series you now have a single spring with half the effective spring constant.

To see this, consider two spring with constants $k_1$ and $k_2$. Since the springs exert identical forces on each other we have $k_1 \Delta x_1 = k_2 \Delta x_2$ where the $\Delta$'s measure the compression or expansion of each spring. Also, $\Delta x_1 + \Delta x_2 = x$ the displacement of the combined spring.

The force exerted by the combined spring is $-k_2 \Delta x_2$ and using the relations above you can see that $F = - \frac {k_1 k_2}{k_1+k_2} x$

 

[robphy]

Let us hang a mass M from spring 1: so, k₁ x₁=Mg.
Let us hang a mass M from spring 2: so, k₂ x₂=Mg.

If we hang the mass M from the springs arranged in series,
we have an effective spring with spring constant K and displacement X=x₁+x₂.
Since KX=Mg, we find
x₁ = KX/k₁ and
x₂ = KX/k₂.

By adding,
X=x₁+x₂=KX(1/k₁ + 1/k₂)
or
K=(1/k₁ + 1/k₂)^-1

When k₁=k=k₂, we have K=k/2.

Edit:
Here's a quick proof using force and energy (instead of displacement).
k₁ x₁²/2+k₂ x₂²/2=KX²/2.
That is,
(k₁ x₁)²/k₁+(k₂ x₂)²/k₂=(KX)²/K.
Since KX=k₁ x₁=k₂ x₂ for springs in series,
(1/k₁+1/k₂)=1/K.

 

[e(ho0n3]

Let us hang a mass M from spring 1: so, k₁ x₁=Mg.
Let us hang a mass M from spring 2: so, k₂ x₂=Mg.

If we hang the mass M from the springs arranged in series,
we have an effective spring with spring constant K and displacement X=x₁+x₂.
Since KX=Mg, we find
x₁ = KX/k₁ and
x₂ = KX/k₂.

What if the springs aren't hanging but are placed horizontally?

Why is this: "The force exerted by the combined spring is -k_2 \Delta x_2" ? Wouldn't the other spring also play a role?