## Chapter 7

## Q. 7.7

A mass *m* is dropped from a height *h* onto a linear spring of constant stiffness *k* and negligible mass. Determine the maximum deflection *δ* of the spring, and compare this value with the static spring deflection \delta_S produced by *m*, See Fig. 7.7. Assume that *m* maintains contact with the spring in its motion following the impact.

## Step-by-Step

## Verified Solution

Since the velocity of *m* is zero at both its initial and terminal states at 1 and 3 in Fig. 7.7, the change in its kinetic energy on the path \mathscr{C} is zero. Because the mass of the spring is negligible , its kinetic energy may be ignored . Moreover, all the forces that act on m are constant, workless, or vary only with the particle’s position on \mathscr{C}. Therefore, the work–energy principle may be applied .

The total work done by the forces acting on *m* from its initial position 1 to its final position 3 on \mathscr{C} is determined by (7.21) in which \mathscr{C}=\mathscr{C}_1\cup \mathscr{C}_2. We note that the instantaneous impulsive force of the spring does no work on *m*. The free body diagrams in Fig. 7.7 show that the force acting on m over \mathscr{C}_1 is F_1 = *mg***i** and over \mathscr{C}_2 is F_2=(mg – kx_2) **i**. Hence,

\mathscr{W}=\int_{\mathscr{C}_1}^{}{F.dx} + \int_{\mathscr{C_2}}^{}{F.dx}=\int_{0}^{h}{mgdx_1} + \int_{0}^{\delta}{(mg – kx_2)dx_2}. (7.42a)

The work–energy equation (7.36) thus yields

\mathscr{W}=mgh + mg \delta – \frac{1}{2}k {\delta}^{2} = \Delta {K}=0, (7.42b)

which determines the following deflection of the spring:

\delta=\frac{mg}{k} + \sqrt{(\frac{mg}{k})^{2} + \frac{2mgh}{k}}. (7.42c)

The static deflection that would result from the weight alone is {\delta}_s=mg/k. Use of this relation in (7.42c) gives

\delta={\delta}_S + \sqrt{{\delta}^{2}_S + 2{\delta}_Sh} \geq 2 {\delta}_S. (7.42d)

This formula shows that *the dynamic deflection δ* *is not less than twice the static deflection {\delta}_S*. In particular, if *m* is released just at the top of the spring so that *h* = 0, then \delta = 2{\delta}_S.

The foregoing solution has illustrated the application of the work–energy equation when the work is calculated by use of the path integrals in (7.42a). Alternatively, we recognize that the work \mathscr{W}_g done by the gravitational force acting over the entire path from 1 to 3 in the direction of the displacement is given by \mathscr{W}_g=mg(h + \delta); the work \mathscr{W}_e done by the elastic spring force acting on *m* over the path from 2 to 3 is \mathscr{W}_e=\frac{1}{2}k{\delta}^{2}; and the impulsive force imposed on *m* at state 2 is workless. Hence, the total work done on *m* is \mathscr{W}=\mathscr{W}_g + \mathscr{W}_e=mg(h + \delta) – \frac{1}{2}k {\delta}^{2}, which is the same as (7.42b) derived above.