Chapter 11, Problem 86GP

To determine

To calculate: The maximum amplitude of oscillations (that allow penny to stay top on block)

Answer to Problem 86GP

The maximum amplitude of oscillations (that allow penny to stay top on block) is $A=\frac{Mg}{k}$ .

Explanation of Solution

Given:

Mass of block $=M$

Stiffness constant of spring $=k$

Mass of penny $=m$

Concept used:

Here a normal force acting upward due to contact of a penny with block and a force is acting in an upward direction which is also positive.

Force on penny $F_{net}=mg-F_n=mg$

Calculation:

From the equation

  $F_{net}=mg-F_n=mg$

  $F_n>0$

  $\begin{array}{l}{a}_{down}=g\\ a_{\max }={\omega }^{2}A=\frac{k}{M+m}\times A\\ M\gg m\end{array}$

So, ignore m

  $\begin{array}{l}{a}_{\max }=\frac{k}{M}\times A=g\\ A=\frac{Mg}{k}\end{array}$

Amplitude which is the maximum of oscillations

Conclusion:

The maximum amplitude of oscillations (that allow the penny to stay top on the block) is $A=\frac{Mg}{k}$ .