Problem 13.18

13.18. a)
From the free body diagram, define and angle $\theta$ that is the angle between the wire and the normal to the wall where the wire touches the wall. Then, from summing the forces in the $\hat{y}$ direction, we find that
$mg = 2T \sin \theta$
Solving for the tension yields
$T = \frac{mg}{2 \sin \theta}$
Now, we need to find the angle $\theta$
If we draw a line up the center of the diagram and bisect the $\frac{3L}{4}$ side, we see that
$\cos \theta = \frac{adjacent}{hypotenuse}$
$\cos \theta = \frac{3L/8}{L/2}$
or that $\cos \theta = \frac{3}{4}$
So, $\theta = 41.4{}^{\circ}$
Now, from our equation for velocity of waves in a string, we find that
$v = \sqrt{\frac{T}{\mu}}$
$v = \sqrt{\frac{mg}{2\mu \sin 41.4{}^{\circ}}}$
$v = 30.4 \sqrt{m}$

b)
From part a), we know that
$v = 30.4 \sqrt{m}$
So, plugging in v = 60 m/s we find that
m = 3.89 kg