Cusps in the evolution of closed strings

[rbwang1225]

Homework Statement

This is problem 7.7 in Zwiebach's book, 2ed ed.
In (b) he want us to show that near the cusp, $y\sim x^{2/3}.$
In (d), Check that the period of the motion of the closed string is $\sigma_1/4c$. How many cusps are formed during a period?

Homework Equations

(b) $\overrightarrow X(t_0,\sigma) = \overrightarrow X_0+\frac{1}{2}(\sigma-\sigma_0)^2\overrightarrow T+\frac{1}{3!}(\sigma-\sigma_0)^3\overrightarrow R \\=\frac{1}{2}(\sigma-\sigma_0)^2 T\hat y+\frac{1}{3!}(\sigma-\sigma_0)^3 R(\cos\theta\hat x+\sin\theta \hat y)$, where $|\overrightarrow T|=T \mbox{ and } |\overrightarrow R|=R$.
(d)$\overrightarrow X(t,\sigma) =\frac{1}{2}[\overrightarrow F(u)+\overrightarrow G(v)] =\frac{\sigma_1}{4\pi}(\sin \frac{2\pi u}{\sigma_1}+\frac{1}{2}\sin\frac{4\pi v}{\sigma_1},-\cos\frac{2\pi u}{\sigma_1},-\frac{1}{2}\cos\frac{4\pi u}{\sigma_1}) =\overrightarrow X(t+T,\sigma)$

The Attempt at a Solution

(b) $y=\frac{1}{2}(\sigma-\sigma_0)^2[T+\frac{1}{3}(\sigma-\sigma_0)R\sin\theta]$
$x=\frac{1}{3!}(\sigma-\sigma_0)^3R\cos\theta$
But, I don't see very clearly why $y\sim x^{2/3}$.
(d) I observe that $\overrightarrow F(u) \mbox{ and } \overrightarrow G(v)$ has periods $\sigma_1/c \mbox{ and }\sigma_1/2c$, respectively. But I don't know why the period of $\overrightarrow X$ is smaller.

Any advice would be very appreciated.

 

[George Jones]

$y=\frac{1}{2}(\sigma-\sigma_0)[T+\frac{1}{3}(\sigma-\sigma_0)^2R\sin\theta]$
$x=\frac{1}{3!}R\cos\theta$
But, I don't see very clearly why $y\sim x^{2/3}$.

These aren't quite correct. Try finding them again. Don't factor out any powers of $\sigma-\sigma_0$.

 

[rbwang1225]

These aren't quite correct. Try finding them again. Don't factor out any powers of $\sigma-\sigma_0$.

I corrected it.
Or, I still have something lost?
Thanks a lot!

 

[George Jones]

It looks like you're still missing something in the expression for y.

 

[rbwang1225]

Oh, sorry for the mistake!
Now, can I conclude that since $y\sim (\sigma-\sigma_0)^2 \mbox{ and } x\sim (\sigma-\sigma_0)^3$ so $y\sim x^{2/3}$?
I think it's yes, because the factors, T, and x-component of R are just numbers, they don't affect the main feature of the line.
Thank you very much!

 

[George Jones]

Roughly, yes.

Maybe better is to solve the $x$ equation for $(\sigma-\sigma_0)$, and substitute this into the $y$ equation, so as to obtain $y$ as a function of $x$. The expression for $y$ should contain two terms that involve $x$. Argue that, near a cusp, one of these terms dominates.

 

[rbwang1225]

OK, then I know how to get the answer.
But now I have a question in (d). I don't know how to derive the period of the function of the sum of 2 periodic functions. I thought it was the smaller one between the two, but it seems like something was missing.