Answer
(a) $v_{max} = 3.93~cm/s$
(b) $a_{max} = 1.54~cm/s^2$
Work Step by Step
(a) From the graph, we can see that one cycle is completed in a period of 16.0 seconds. We can find the angular frequency.
$T = \frac{2\pi}{\omega}$
$\omega = \frac{2\pi}{T}$
$\omega = \frac{2\pi}{16.0~s}$
$\omega = 0.3927~rad/s$
From the graph, we can see that the amplitude is 10.0 cm. We can find the maximum speed.
$v_{max} = A~\omega$
$v_{max} = (10.0~cm)(0.3927~rad/s)$
$v_{max} = 3.93~cm/s$
(b) We can find the maximum acceleration.
$a_{max} = A~\omega^2$
$a_{max} = (10.0~cm)(0.3927~rad/s)^2$
$a_{max} = 1.54~cm/s^2$
