# Thread: [SOLVED] Endless Chord - Oscillation?

1. ## [SOLVED] Endless Chord - Oscillation?

An endless chord consists of two portions of lengths $2l$ and $2l'$ respectively, knotted together. The mass per unit length of each string is $m$ and $m'$. It is placed in stable equilibrium over a smooth peg and slightly displaced. Find time period of oscillation.

Spoiler:
$T = 2\pi\sqrt {\frac {ml + m'l'}{(m' - m)g}}$

Any ideas?

2. This one is actually quite easy!

Here's how to solve it but I must point out that this can only be true if $\color[rgb]{1,0,0} m' > m$ .

Spoiler:

OK, so the mechanics of this problem are that the two chords have different densities and so when displaced this results in a mass difference on either side. If the heavier chord is at the bottom this will result in SHM as the direction of the net weight weight will be restoring in nature.

Now we can assume that the heavier chord is at the bottom since if it were at the top displacement from equilibrium would cause it to fall to the bottom (also resulting in oscillations). Hence we only need consider the case where the heavier chord is at the bottom.

First consider what constitutes equilibrium here. This will be when each side is symmetric i.e. equal lengths of each chord must be either side of the peg and so if we call their joining knots $A$ and $B$ then $A$ and $B$ must be at the same height.

Now if we consider the displacement of $A$ upwards by a distance $x$ from the equilibrium point then for the region of the chords greater than $x$ above and below from the equilibrium point there is symmetry and so the masses balance. However, within a region $x$ above and below the equilibrium point the masses do not balance providing net force

$F = 2 x (m' - m) g$

which is directed downwards at $A$.

This force is equally transmitted through the chords and so the inertial mass acted upon by this force is the total mass of the chords so:

$-2 x (m' - m ) g = 2 (m' l' + m l) \ddot{x}$ ,

$\iff \ddot{x} = - \frac{(m'-m)g}{m' l' + ml} x$.

Hence

$\omega^2 = \frac{(m'-m)g}{m' l' + ml}$

and as $\omega T = 2 \pi$

$\color[rgb]{0,0,1} \boxed{T = 2 \pi \sqrt{\frac{m' l' + ml}{(m'-m)g}} }$