Answer
The transverse speed for the point is $~~133~cm/s$
Work Step by Step
$y(x,t) = 15.0~sin(5\pi x/8 - 4 \pi t)$
We can take the derivative to find an equation for the transverse velocity:
$u = (15.0)(-4\pi)~cos(5\pi x/8 - 4 \pi t)$
$u = (-60\pi)~cos(5\pi x/8 - 4 \pi t)$
We can find the transverse velocity for the point:
$u = (-60\pi)~cos(5\pi x/8 - 4 \pi t)$
$u = (-60\pi)~cos[(5\pi) (6.00~cm)/8 - (4 \pi) (0.250~s)]$
$u = (-60\pi)~cos (2.75\pi)$
$u = 133~cm/s$
The transverse speed for the point is $~~133~cm/s$
