# Harmonic Motion [closed]

A light elastic string is stretched between two points, one lying vertically below the other. A particle is attached to the midpoint 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 is $$2\pi(h/g)^{1/2}$$

Can someone tell me how to tackle this question? I have a general formula for an approximate period of a small oscillation about an equilibrium point in my notes. I think I have to calculate the potential of the particle, this is what's tripping me up. Can someone help me?

## closed as off-topic by tpg2114♦, Emilio Pisanty, David Z♦Nov 5 '13 at 17:19

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The elastic string is behaving like a spring, and for a spring we know the restoring force when you stretch it a distance $x$ is $F = -kx$ where $k$ is the spring constant. In this case we aren't told the force constant, but we know that when you attach a particle of some mass $m$ the cord stretches by $h$, so:

$$mg = -kh$$

or:

$$k = - \frac{mg}{h}$$

So if you displace your particle by a distance $x$ the restoring force, $F(x)$, is:

$$F(x) = - \frac{mg}{h}x$$

and since $F = ma$ we divide by $m$ to get $a$:

$$a = \frac{d^2x}{dt^2} = - \frac{g}{h}x$$

Now solve this differential equation and from the equation of motion you can calculate the period.

I assume you have drawn a free body diagram of the mass (give it a mass $m$; it won't matter) and identified the forces acting on it using Hooke's law. You can label the particle's position by it's $y$ from the top, coordinate if you like. Then, into $F_{net} = m a$ everything goes. Compare the result with some of the force equations of other simple harmonic problems you have encountered, and work from there. Remember, few of the things you learn are in isolation.