# M07M3

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M07M3 is a short name for the third problem in the Classical Mechanics section of the May 2007 Princeton University Prelims. The problem statement can be found in the problems list. Here is the solution.

(a)

Define x as the coordinate that goes downward and y as the coordinate that goes to the left. Take a small piece of string and write its horizontal and vertical equations of motion:

$\tau cos\theta(x)=\tau cos(\theta(x+dx))+\frac{m}{l}dx g$

$\tau sin(\theta(x))\theta'(x)=\frac{m}{l} g \quad Vertical$

$\frac{m}{l}dx\ddot{y}=\tau sin[\theta(x)+\theta'(x)dx]-\tau sin\theta(x)=\tau cos\theta(x) \theta'(x) dx$

$\frac{m\ddot{y}}{l}=\tau cos\theta \theta' \quad Horizontal$

In all these equations, $\theta$ is defined as the angle that a differential piece of string makes with the vertical line. We can express it in terms of y as:

$tan\theta=y'$

$sec^2\theta \theta'=y''$

Plug this in to get:

$\tau sin\theta y'' cos^2\theta=\frac{mg}{l}$

$\frac{m\ddot{y}}{l}=\tau y'' cos^3\theta$

At the top we have y=0, and at the bottom:

$\tau cos\theta(l)=Mg$

$-\tau sin\theta(l)=M\ddot{y}(l)$

For small oscillations, these become:

$\frac{m\ddot{y}}{l}=\tau y''$

$\tau=Mg$

Failed to parse (unknown function\label): \label{bottom}-\tau y'(l)=M\ddot{y}(l)

Failed to parse (unknown function\label): \label{top}y(0)=0

(b)

Let $v^2 \equiv \frac{\tau l}{m}$, and let $k\equiv\omega/v$. Then a wavemode with frequency $\omega$ has a form:

$y(x,t)=Asin(\omega t+\phi)sin(kx+\varphi)$

Failed to parse (unknown function\ref): y(0,t)=Asin(\omega t+\phi)sin\varphi=0 \rightarrow \varphi=0 \quad (\ref{top})

Failed to parse (unknown function\ref): kgcos(kl)=\omega^2sin(kl) \rightarrow (kl) tan(kl)=\frac{m}{M} \quad (\ref{bottom})

$(\omega l/v)tan(\omega l/v)=\frac{m}{M}$

(c)

To lowest order we get $\omega=0$, but then there is no motion. To first order in $m/M$, we get:

$(kl)^2=\frac{m}{M}$

$\omega=\sqrt{\frac{g}{l}}$

The whole system swings like a pendulum of length l. The next lowest frequency will be for the case when (kl) is not small but tan(kl) is:

$kl \approx \pi +\epsilon$

$\omega=\pi\sqrt{\frac{Mg}{ml}}$

In this case, the point mass remains fixed, acting as a node, while the string oscillates back and forth. You can see that the frequency is much larger than the one for pendulum motion.