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A light elastic string is stretched between two points, one lying vertically below the other. A particle is attached to the mid-point of the string, causing it to sink a distance h. Assuming that the string below the particle does not go slack, show that the period of small vertical oscillations about its equilibrium position is $2\pi(h/g)^{1/2}$.

My attempt:

The string seems to be acting like a spring so from Hess's law we have the tension $T = kh$ and resolving downwards we have $mg = kh$ so $k = \frac{mg}{h} $, this was as far as I was able to go - how else would I proceed? I've tried using $F = ma$ but no luck.

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marked as duplicate by John Rennie, Dimensio1n0, tpg2114, Qmechanic Nov 10 '13 at 19:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Given small vertical oscillations, you can assume the motion to be simple harmonic. Then the time period would be given by $2\pi/\omega$. –  udiboy1209 Nov 10 '13 at 13:55
I have checked the other thread - that's the exact same question but I still don't understand the responses there. What does "So if you displace your particle by a distance x the restoring force" how did he/she get the following equation from there? The current methods which we do in class to find the simple harmonic oscillator is to find the potential V(X), then find the potion X where the particle is in equilibrium (no idea what it is in this case) then use $2\pi/w$ with V''(X) –  user33484 Nov 10 '13 at 14:19

1 Answer 1

You need to take your attempt a little further, rather than just resolving the various forces: Try setting up a differential equation for your problem, you know that the particle is oscillating so you can propose a trial solution to your equation (something along the lines of $y=Ae^{ix}$)

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We haven't gone over differential equations yet –  user33484 Nov 10 '13 at 13:59

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