Problem 12.46

12.46. a)
The displacement of each spring is given by x₁, and x₂.
So the total displacement of the block is given by
x = x₁+x₂
So, at the point between the two springs, the forces are equal, so
k₁x₁ = k₂x₂
substituting in for x₂, we find
k₁x₁ = k₂(x - x₁)
Solving for x₁, we find
$x_{1} = \frac{k_{2}x}{k_{1} + k_{2}}$
and the force on spring 1 is
$F_{1} = \frac{k_{1}k_{2}x}{(k_{1}+k_{2})} ~=~ ma$
This is of the form
F = k_eff x  =  ma
and the period of the motion is given by
$T = 2\pi \sqrt{\frac{m}{k_{eff}}}$
or $T = 2 \pi \sqrt{\frac{m(k_{1}+k_{2})}{k_{1}k_{2}}}$

b)
Both springs are stretched by a distance x, so summing the forces gives
F = -(k₁+k₂)x
k_eff = (k₁+k₂)
so solving for the period, we find,
$T = 2\pi \sqrt{\frac{m}{(k_{1}+k_{2})}}$