How to approximate this simple harmonic motion?

Question

In the assembly shown in the figure, both the pullies are massless and frictionless on their axes. The positions of the strings are adjusted in such a way that, in equilibrium position, springs $k_1$ and $k_2$ are unstretched. Now if $m$ is given a very slight displacement in the vertical direction, what will be its time period for small vertical oscillations (in seconds)? Given that $k_1 = k_2 = k_3 = k = 6\pi^2\,\mathrm{N\,m^{-1}}$. The string that connects the ends of $k_1$ and $k_2$ is inextensible, and is wrapped around the pulley $P_1$ to ensure no slipping. The string that connects the mass and $k_3$ passes over both pullies and does not slip over them. (Take the mass of the block $m=1\,\mathrm{kg}$)

(A) $\sqrt{3/2}$
(B) $\sqrt 2/3$
(C) $1/\sqrt 3$
(D) $\infty$

Answer: (B) $1/\sqrt 3$

How do I approach this question? I really have no clue.

In particular, I do not understand how to account for the effect of the $k_2$ and $k_3$. Since they are not directly connected to the string attached to the mass $m$, so it becomes harder to find the relation. Let us say we displace the block by a distance $x$ downwards from its equilibrium position (we can effectively ignore gravity). $k_1$ and $k_3$ stretch by a length $x$, and for this question, we will assume $k_2$ to compress by a length $x$ (I know it is incorrect, but without that assumption, I think this question will become very complicated).

So how do these $k_1$ and $k_2$ affect the tension in the string? First I assumed that they had no effect, but that gave an answer of $\sqrt{2/3}$ seconds. So then I assumed that they will add together, so that the tension force on the block would be $T=(k_1 + k_2 + k_3)x$, which gave an answer of $\sqrt 2/3$ seconds. But that is incorrect! And what's more, it isn't clear to me how my expression for $T$ is incorrect or correct.

So what is the correct answer? And even if the correct answer is (B), what is the conceptual explanation behind it?

Thank you.

Answer 1

When $k_1$ and $k_3$ are in elongation, $k_2$ is in compression, so the equation of motion of the mass is:

$$m\ddot{y}=-k_3y-k_1y-k_2y+mg$$ $$\ddot{y}+\frac{(k_3+k_1+k_2)}{m}y=g$$ This has solutions of the type:

$$y=A\cos\omega t$$ Where: $$\omega^2=\frac{(k_3+k_1+k_2)}{m}$$ $$\omega=\frac{2\pi}{T}\implies T=\frac{2\pi}{\omega}$$ $$\implies T=2\pi\sqrt{\frac{m}{(k_3+k_1+k_2)}}$$ $$T=\frac{2\pi}{\sqrt{18\pi^2}}=\frac{2}{\sqrt{2\times9}}=\frac{\sqrt{2}}{3}$$ The answer is $B$.

---

Edit: (answering OPs question in the comment section)

Initially (left) none of the springs are extended or compressed.

Then (right) we introduce a small perturbation by turning $P_1$ slightly counterclockwise, see the angle $\theta$. The displacement of $m$, is: $$y=R\theta$$ With $R$ the radius of $P_1$.

Now note that the bits connecting $P_1$ to $k_1$ and $k_2$ and the bits connecting $k_1$ and $k_2$ to the wall are all rigid. The point $A$ has moved to $A'$ and $B$ to $B'$.

Again the displacements are:

$$R\theta$$

Which leads to the equation of motion given higher up.