Question 1.1: The sinusoidal vibration displacement amplitude at a particu......

Remembering Eq. (1.1),

x=X \sin \omega t                                                               (A)

we simply differentiate twice, so,

\dot{x} = \omega X \cos \omega t                                                           (B)

and

\ddot{x} =- \omega^{2} X \sin \omega t                                                       (C)

The single-peak displacement, X, is, in this case, 1.00 mm or 0.001 m. The value of \omega =2\pi f , where \mathcal{f} is the frequency in Hz. Thus, ω = 2π(20) = 40π rad/s.

From Eq. (B), the single-peak value of \dot{x} is ωX, or (40π × 0.001) = 0.126 m/s or 126 mm/s.
From Eq. (C), the single-peak value of \ddot{x} is ω²X or [(40π)²× 0.001]=15.8 m/s² or (15.8/9.81) = 1.61 g.
Root mean square values are {1}/{\sqrt{2} } or 0.707 times single-peak values in all cases, as shown in the Table 1.1.