Answer
The second time that the particle at $x = 0$ has zero velocity is $~~t = 0.50~s$
Work Step by Step
We can use superposition to find the equation for the standing wave:
$y'(x,t) = (0.050)~cos~(\pi x-4\pi t) + (0.050)~cos~(\pi x + 4\pi t)$
$y'(x,t) = (2)(0.050)~cos~(\pi x)~cos~(4\pi t)$
$y'(x,t) = (0.10)~cos~(\pi x)~cos~(4\pi t)$
The particle at $x = 0$ has zero velocity when the point has the maximum transverse displacement of $~~y = y_m = 0.10~m~~$ or $~~y = -y_m = -0.10~m$
We can find values of $t$, where $t \geq 0$, such that $y'(0,t) = 0.10$ or $y'(0,t) = -0.10$:
$\vert y'(0,t) \vert = \vert (0.10)~cos~[(\pi)(0)]~cos~(4\pi t) \vert = 0.10$
$\vert (0.10)~cos~(0)~cos~(4\pi t) \vert = 0.10$
$\vert cos~(4\pi t) \vert = 1$
$4\pi t = 0, \pi, 2\pi, 3\pi,...$
$t = 0, 0.25~s, 0.50~s, 0.75~s,...$
The second time that the particle at $x = 0$ has zero velocity is $~~t = 0.50~s$
