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Mechanical Oscillations

Problem 4.35

A plank with a body of mass mm placed on it starts moving straight up according to the law y=a(1−cos⁡ωt),y=a(1-\cos \omega t), where yy is the displacement from the initial position, ω=\omega= =11 s−1.=11 \mathrm{~s}^{-1} . Find: (a) the time dependence of the force that the body exerts on the plank if a=4.0 cm;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 cmh=50 \mathrm{~cm} relative to the initial position (at the moment t=0t=0 ).

Reveal Answer
 4.35.  (a) F=mg(1++aω2gcos⁡ωt), see Fig. 30; (b) amin⁡=g/ω2=8 cm; (c) a==(ω2h/g−1)g/ω2=20 cm\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}
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