Problem 4.35

A plank with a body of mass \(m\) placed on it starts moving straight up according to the law \(y=a(1-\cos \omega t),\) where \(y\) is the displacement from the initial position, \(\omega=\) \(=11 \mathrm{~s}^{-1} .\) Find: (a) the time dependence of the force that the body exerts on the plank if \(a=4.0 \mathrm{~cm} ;\) plot this dependence; (b) the minimum amplitude of oscillation of the plank at which the body starts falling behind the plank; (c) the amplitude of oscillation of the plank at which the body springs up to a height \(h=50 \mathrm{~cm}\) relative to the initial position (at the moment \(t=0\) ).

\begin{aligned} &\text { 4.35. } \quad \text { (a) } \quad F=m g \quad(1+\\ &\begin{array}{l} \left.+\frac{a \omega^{2}}{g} \cos \omega t\right), \text { see Fig. } 30 ; \\ \text { (b) } a_{\min }=g / \omega^{2}=8 \mathrm{~cm} ; \text { (c) } a= \\ =(\omega \sqrt{2 h / g}-1) g / \omega^{2}=20 \mathrm{~cm} \end{array} \end{aligned}

Mechanical Oscillations