Problem 13.7

13.7. a)
The amplitude is 8 cm = 0.08 m.
The wave number is given by
$k = \frac{2\pi}{\lambda}$
k = 7.85 m^-1
$\omega = 2\pi f$
$\omega = 6 \pi rad/s$
Putting it all together, we have
$y(x,t) = A \sin (kx + \omega t + \phi)$
$y(x,t) = (0.08 m) \sin (7.85 x + 6 \pi t)$
where $\phi = 0$ because y = 0 when t = 0.

b)
In general,
$y(x,t) = A \sin (kx + \omega t + \phi)$
Assuming that
y(.1,0) = 0
Then
$0 = .08 \sin (.785 + \phi)$
or, $\phi = -.785$
Therefore,
$y(x,t) = .08 \sin (7.85x + 6\pi t - .785) m$