# Why the resultant spring constant different in the following two cases?

In these two cases in the first case my book The Physics Of Waves And Oscillations by NK Bajaj says:

That the restoring force exerted on the mass by each spring is same so added the elongations and he got k =$\frac{k_1 k_2}{k_1 +k_2}$ (for case a). And in the second case the total restoring force is $-k_2 x -k_1 x =F$ so $k'x = -(k_2 +k_1) x$ so $k' =k_1 + k_2$ (case b).

Now if I say in the first case (case a) the while the spring is released from its stretched position then the spring attached to mass wants to restore it in its equilibrium and so does the second. So shouldn't the attached spring pull the mass simultaneously being pulled by the another spring. In this pursuit the mass also gets double acceleration so the restoring force should be $k_1x_1 +k_2x_2$ instead F. So the question is if this is true in the both cases we should get the same restoring force. Where am I wrong?

• note that the difference between these 2 cases is exactly "adding in series" vs "adding parallel". So under which case do the restoring forces add linearly? Then in the other ("more complicated") case, they will add harmonically. – JEB Sep 17 '18 at 13:25
• Did you notice the rigid constraint on the right hand side of figure b, and the absence of constraint on the right hand side of figure a? – Chet Miller Sep 17 '18 at 13:51

## 2 Answers

As you say case (b) is straightforward, but case (a) need a bit of thought.

Suppose we displace the mass by a distance $x$. This means we stretch both the springs. Let's call the stretch of the first spring $x_1$ and the stretch of the second spring $x_2$. Together these must add up to the total stretch, so we have:

$$x_1 + x_2 = x \tag{1}$$

The tension in the first spring is $k_1 x_1$ and the tension in the second spring is $k_2 x_2$, and these must be equal for the springs to be in equilibrium with each other. So we get a second equation:

$$k_1 x_1 = k_2 x_2 \tag{2}$$

So we have a pair of simultaneous equations for $x_1$ and $x_2$. If we use equation (1) to write:

$$x_1 = x - x_2$$

And substitute for $x_1$ in equation (2) we get:

$$k_1 ( x - x_2) = k_2 x_2$$

and this rearranges to give us $x_2$:

$$x_2 = \frac{k_1}{k_1 + k_2} x$$

So the tension in spring 2, $k_2 x_2$ is:

$$F_2 = \frac{k_1 k_2}{k_1 + k_2} x$$

This is equal to the tension in spring 1, and both are equal to the force on the mass. So we find the combined spring constant is:

$$k = \frac{k_1 k_2}{k_1 + k_2}$$

Consider what happens when the mass moves a distance $x$

In (b) both springs suffer an extension/contraction equal to $x$ and produce different magnitude forces on the mass $k_1x$ and $k_2x$ which are in the same direction.
The net force on the mass is $(k_1+k_2)x$ which leads to an effective spring constant of $k_1+k_2$.

In (a) each spring suffers extensions $x_1$ and $x_2$ which are smaller than $x$ and must add up to be equal to $x$.
The forces exerted by each spring will be in the same direction and of the same magnitude $F$.
In this configuration the combined spring is “weaker” than each of the individual springs.

So $x=x_1+x_2 \Rightarrow \dfrac{F}{k_{\rm effective}}=\dfrac{F}{k_1}+ \dfrac{F}{k_2}$ and the result follows.