Page:CarmichealMass.djvu/7

and therefore the potential energy^[1] is proportional to θ², say that it is ${\tfrac {1}{2}}k\theta ^{2}$. Now from the values of y and x above we have

$y=x{\sqrt {1-\beta ^{2}}}\tan \theta$

For small oscillations we have x = 1 and tan θ = θ; and therefore

$y={\sqrt {1-\beta ^{2}}}\cdot \theta$

Hence the potential energy is

${\frac {1}{2}}{\frac {k}{1-\beta ^{2}}}y^{2}$;

and the equation of motion of the particle becomes

$m_{1}{\frac {d^{2}y}{dt^{2}}}=-{\frac {k}{1-\beta ^{2}}}y$

Hence the period $T_{1}$ of oscillation is

$T_{1}=2\pi {\sqrt {\frac {m_{1}\left(1-\beta ^{2}\right)}{k}}}$

In the second case — when the rod is parallel to the line of relative motion of <marh>S_1</math> and $S_{2}$ — the amount of twisting in the wire for a given position of the balls is the absolute value of $(\pi /2)-\theta$. The potential energy is ${\tfrac {1}{2}}k\left[(\pi /2)-\theta \right]^{2}$. We have

$x={\frac {y}{\sqrt {1-\beta ^{2}}}}\cot \theta$

For small oscillations we have

$y={\sqrt {1-\beta ^{2}}},\ \cot \theta =\tan \left({\frac {\pi }{2}}-\theta \right)={\frac {\pi }{2}}-\theta .$

Hence the potential energy is ${\tfrac {1}{2}}kx^{2}$ the period $T_{2}$ of oscillation is therefore

$T_{2}=2\pi {\sqrt {\frac {m_{2}}{k}}}.$

Equating the two periods of oscillation found above we have

$m_{2}=\left(1-\beta ^{2}\right)m_{1}.$

1. That the potential energy is proportional to $\theta ^{2}$ when measured by B is obvious. Since A observes a different apparent angle $\theta '$ (say) corresponding to $B$'s observed angle $\theta$ it might at first sight appear that the potential energy as observed by A is proportional to $\theta '^{2}$. That this is not the case is seen from the fact that for a given twist in the wire $\theta '$ depends on the direction of equilibrium of the bar, that is, it depends on the way in which the bar is attached to the wire; hence, if the potential energy as observed by A were proportional to $\theta '^{2}$ it would depend on the way in which the bar is attached. Since this is obviously not the case we conclude that the potential energy is proportional to $\theta ^{2}$.