Answer
The transverse speed of a particle of the string at the position $x = 1.5 ~cm$ when $t = \frac{9}{8} s$ is zero.
Work Step by Step
We can write a general expression for the standing wave:
$y'(x,t) = [(2y_m) sin ~(kx)]~cos~\omega t$
We can write a general expression for the transverse speed at a point $x$:
$u = -\omega~[(2y_m) sin ~(kx)]~sin~\omega t$
We can find the transverse speed of a particle of the string at the position $x = 1.5 ~cm$ when $t = \frac{9}{8} s$:
$u = -\omega~[(2y_m) sin ~(kx)]~sin~\omega t$
$u = -(40\pi s^{-1})~[(0.50~cm) sin ~(\frac{\pi}{3}cm^{-1} \cdot 1.5~cm)]~sin~[(40\pi s^{-1})(\frac{9}{8} s)]$
$u = -(40\pi s^{-1})~[(0.50~cm) sin ~(\frac{\pi}{2})]~sin~(45\pi)$
$u = -(40\pi s^{-1})~(0.50~cm)~(0)$
$u = 0$
The transverse speed of a particle of the string at the position $x = 1.5 ~cm$ when $t = \frac{9}{8} s$ is zero.
