Effective spring constant of springs in series When we consider the analogy of a rod as a spring while considering the combination of rods for finding the stretch we do this-
Let for any rod considered as a spring 
$$k=\frac{AY}{L}$$
If two rods are connected and undergo stretching due toa force. 
For ex: Two connected rods hanging in series on a ceiling.
The effective spring constant $k$ is found by
$$\frac{1}{k1}+\frac{1}{k2}=\frac{1}{k}$$
Can you give me a proof for this.
 A: Let there be a block of mass M placed on a smooth horizontal surface connected to a wall by a wire consisting of two springs in series of spring constants, say, $k_1$ and $k_2$ respectively.

If I pull the block towards right by x units then let the stretch in spring 1 and 2 be $x_1$ and $x_2$ respectively.
Therefore, $x=x_1 + x_2 \tag{1}$
And the restoring forces in the springs will be given by,
$F_1=-k_1 x_1$
$F_2=-k_2 x_2$
Let the net force be $F$ and $F=-kx$.
From equation $(1)$,
$-\frac{F}{k}=-\frac{F_1}{k_1} - \frac{F_2}{k_2} \tag{2}$ 
But the force in the wire will be the same because it is massless and ideal.The force that each spring experiences will have to be same, otherwise the springs would buckle.  Thus, $F=F_1=F_2$.
On canceling the force terms in equation $(2)$, we get
$\frac{1}{k}=\frac{1}{k_1} + \frac{1}{k_2}$
If two springs were connected parallel then the net force would be give by $F=F_1+F_2$, and on solving further by putting $F=-kx$ , $F_1=-k_1 x_1$ and $F_2=-k_2 x_2$ you would get $k=k_1 +k_2$.
There will be a similar proof in the case of a block hanged by two springs to a ceiling.
