Answer
The maximum speed of the needle is $7.0~cm/s$
Work Step by Step
We can find the period:
$T = \frac{9.0~s}{24~cycles} = 0.375~s$
We can find $\omega$:
$T = \frac{2\pi}{\omega}$
$\omega = \frac{2\pi}{T}$
$\omega = \frac{2\pi}{0.375~s}$
$\omega = 16.755~rad/s$
The amplitude is half the distance between the lowest and highest point of the motion. We can find the amplitude:
$A = \frac{8.4~mm}{2} = 4.2~mm$
We can find the maximum speed of the needle:
$v_m = A~\omega$
$v_m = (4.2~mm)~(16.755~rad/s)$
$v_m = 70~mm/s$
$v_m = 7.0~cm/s$
The maximum speed of the needle is $7.0~cm/s$
