Answer
The magnitude of the transverse acceleration for the point is $~~16.7~m/s^2$
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 take the derivative to find an equation for the transverse acceleration:
$a = -(-60\pi)(-4\pi)~sin(5\pi x/8 - 4 \pi t)$
$a = (-240\pi^2)~sin(5\pi x/8 - 4 \pi t)$
We can find the transverse acceleration for the point:
$a = (-240\pi^2)~sin(5\pi x/8 - 4 \pi t)$
$a = (-240\pi^2)~sin[(5\pi) (6.00~cm)/8 - (4 \pi) (0.250~s)]$
$a = (-240\pi^2)~sin (2.75\pi)$
$a = -1670~cm/s^2$
$a = -16.7~m/s^2$
The magnitude of the transverse acceleration for the point is $~~16.7~m/s^2$
