Answer
The magnitude of the maximum transverse acceleration for any point is $~~23.7~m/s^2$
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 magnitude of the maximum transverse acceleration for any point:
$a = (-240\pi^2)~sin(5\pi x/8 - 4 \pi t)$
$a = (-240\pi^2)~(-1)$
$a = 240\pi^2$
$a = 2370~cm/s^2$
$a = 23.7~m/s^2$
The magnitude of the maximum transverse acceleration for any point is $~~23.7~m/s^2$
