# How to approximate this simple harmonic motion?

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.

• Also, I would love some tips to solve general SHM approximation questions, along with how to draw these diagrams on PC, better. Many answerers here have diagrams worthy of posting in electoral campaigns! – FreezingFire Oct 7 '16 at 13:19
• You can always ask someone as a comment to their answers, if you wish some info about their tools ☺️ – Steeven Oct 7 '16 at 13:42
• So is (B) $\sqrt 2/3$ or $1/\sqrt 3$? – JEB Feb 5 at 14:15

## 1 Answer

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.

• Thank you for your answer! I used the same procedure, what I did not understand is why the forces of the three springs are in simple addition! I mean, how do $k_1$ and $k_2$ affect the block's motion? – FreezingFire Oct 7 '16 at 15:27
• $k_3$ acts like a normal spring: it's as if the mass is suspended from the ceiling by that spring. When the mass moves a little downwards, $P_1$ turns slightly counterclockwise, causing $k_1$ to extend a bit and $k_3$ to compress a bit. The net result is that you could replace $k_3$ by $k_3+k_2+k_1$ and remove $k_1$ and $k_2$. The extension/compression of all springs is always $y$, due to the simple geometry of the situation. – Gert Oct 7 '16 at 15:48
• But, the springs $k_1$ and $k_2$ apply a force on the pulley, so how is it transmitted to the rope? For this to happen, the tension in the two parts of the rope must be different. But neither the pulley has inertia, nor the rope has any mass. So what is the reason behind this transmission of the extra force to the rope? – FreezingFire Oct 7 '16 at 17:29
• I'll see if I can come up with a sketch. – Gert Oct 7 '16 at 17:31