Oscillating a spring

Relevant wiki: Hooke's Law

At the point P there are 2 Forces (at t=0): ${ k }_{ 1 }{ A }_{ 1 }$ and ${ k }_{ 2 }{ A }_{ 2 }={ k }_{ 2 }{ (A- }{ A }_{ 1 })$ , because springs are connected consistently.

${ k }_{ 1 }{ A }_{ 1 }={ k }_{ 2 }{ (A- }{ A }_{ 1 })\\ ({ k }_{ 1 }+{ k }_{ 2 }){ A }_{ 1 }={ k }_{ 2 }A\\ { A }_{ 1 }=\frac { { k }_{ 2 } }{ ({ k }_{ 1 }+{ k }_{ 2 }) } A$

I liked the Lagrangian magic of Steven Chase's solution. But suppose you were in an exam situation and didn't have time to do the detailed calculation?

I reasoned physically as follows. Suppose the two springs had the same spring constant. Then the point between them will oscillate with half the amplitude of the block. So we know the answer must be option four or five!

Now suppose we make $k_2 >> k_1$ so big that the spring attached to the block hardly stretches at all during the motion. Then P has almost as big an amplitude as the block, allowing us to identify the fifth option as the correct one.

This is a good opportunity to practice Lagrangian Mechanics . Define $s_1$ and $s_2$ as the stretches of the two springs.

Spring Potential Energy: $U = \frac{1}{2} k_1 s_1^2 + \frac{1}{2} k_2 s_2^2$

Mass Kinetic Energy: $E = \frac{1}{2} m (\dot{s_1} + \dot{s_2})^2$

Lagrangian:

$L = E - U = \frac{1}{2} m (\dot{s_1} + \dot{s_2})^2 - \frac{1}{2} k_1 s_1^2 - \frac{1}{2} k_2 s_2^2$

Equations of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{s_1}}} = \frac{\partial{L}}{\partial{s_1}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{s_2}}} = \frac{\partial{L}}{\partial{s_2}}$

Evaluating these results in:

$m (\ddot{s_1} + \ddot{s_2}) = -k_1 s_1 \\ m (\ddot{s_1} + \ddot{s_2}) = -k_2 s_2 \\ \implies k_1 s_1 = k_2 s_2$

Mass displacement (total stretch) in terms of spring stretches:

$s = s_1 + s_2 = s_1 + \frac{k_1}{k_2} s_1 = s_1 \frac{k_1 + k_2}{k_2} \\ \implies \boxed{s_1 = s \frac{k_2}{k_1 + k_2}}$

The force is always equal at two important points. One where the mass is attached to spring (of spring constant k2) and the point P acting on spring constant (k1). This is true since a force applied to spring with spring constant k2 will ultimately have to exert the same force on spring with spring constant k1 as they are always in tension. The maximum extension of spring with spring constant k1 = to the amplitude of point P. This maximum extension is created when the force acting on spring k1 with maximum force from hooke's law (F=Ke). We also know that the force on spring k1 is the same as on spring k2 hence their maximum extensions occur at the same time. The maximum extension of both springs added together is the extension of the spring. Hence from this information: F=k1 e1=k2 e2 e1+e2=A Given two different equations involving both e1 and e2 they can be simultaneously solved for e1 (the maximum extension of spring 1). Which is the amplitude of point P.

Assume spring K1 extremely hard and K2 extremely soft. K1 will not move almost, K2 will squeeze completely. This suggests the only possible solution - the last listed.