Chapter 19.1, Problem 19.10P

A 5-kg fragile glass vase is surrounded by packing material in a cardboard box of negligible weight. The packing material has negligible damping and a force-deflection relationship as shown. Knowing that the box is dropped from a height of 1 m and the impact with the ground is perfectly plastic, determine (a) the amplitude of vibration for the vase, (b) the maximum acceleration the vase experiences in g' s.
  

To determine

(a)

The amplitude of vibration for the vase.

Answer to Problem 19.10P

Amplitude $x_m=99mm.$

Explanation of Solution

Given information:

Mass of vase $m=5{\rm K}g$

Height $h=1m$

Velocity at the end of free fall $v=\sqrt{2gh}$

$\begin{array}{l}v=\sqrt{2\times 9.81\times 1}\\ v=\sqrt{19.62}\\ v=4.43m/s\end{array}$

Assume that the spring is unstretched during the free fall. To better understand we use a simple spring mass model for the motion of the vase and the packing material.

$m=5{\rm K}g$

Now, taking slope from the graph.

$\begin{array}{l}k=\frac{100}{10}\\ k=10{\rm N}/mmor,\\ k=10000{\rm N}/m\end{array}$

Calculation:

Now, consider simple harmonic motion:

$x=x_m\sin \left({\omega }_{n}t+\varphi \right)$

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

$\begin{array}{l}v=\frac{d}{dt}\left(x_m\sin \left({\omega }_{n}t+\varphi \right)\right)\\ v={\omega }_n{x}_m\cos \left({\omega }_{n}t+\varphi \right)\end{array}$

When the box hits the ground, let $t=0$

Then, at $t=0$

$x=0$

And, velocity, $v={\omega }_n{x}_m$

$\left(\therefore \cos 0^{°}=1\right)$

So, velocity of the vase = velocity at the end of free fall

$4.43={\omega }_n{x}_m$

Now, Natural frequency: ${\omega }_n=\sqrt{\frac{k}{m}}$

$\begin{array}{l}{\omega }_n=\sqrt{\frac{10000}{5}}\\ {\omega }_n=\sqrt{2000}\\ {\omega }_n=44.721rad/s\end{array}$

Thus, amplitude of the vase $x_m$ ,

$\begin{array}{l}4.43={\omega }_n{x}_m\\ x_m=\frac{4.43}{{\omega }_n}\\ x_m=\frac{4.43}{44.721}\\ x_m=0.09905mor,\\ x_m=99mm\end{array}$

Conclusion:

The amplitude of the vase is $x_m=99mm.$

To determine

(b)

The maximum acceleration the vase experiences.

Answer to Problem 19.10P

Acceleration $a=20.2g$

Explanation of Solution

Given information:

Mass of vase $m=5{\rm K}g$

Height $h=1m$

Velocity at the end of free fall $v=\sqrt{2gh}$

$\begin{array}{l}v=\sqrt{2\times 9.81\times 1}\\ v=\sqrt{19.62}\\ v=4.43m/s\end{array}$

Assume that the spring is unstretched during the free fall. To better understand we use a simple spring mass model for the motion of the vase and the packing material.

$m=5{\rm K}g$

Now, taking slope from the graph.

$\begin{array}{l}k=\frac{100}{10}\\ k=10{\rm N}/mmor,\\ k=10000{\rm N}/m\end{array}$

Calculation:

Now, consider simple harmonic motion:

$x=x_m\sin \left({\omega }_{n}t+\varphi \right)$

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

$\begin{array}{l}v=\frac{d}{dt}\left(x_m\sin \left({\omega }_{n}t+\varphi \right)\right)\\ v={\omega }_n{x}_m\cos \left({\omega }_{n}t+\varphi \right)\end{array}$

When the box hits the ground, let $t=0$

Then, at $t=0$

$x=0$

And, velocity, $v={\omega }_n{x}_m$

$\left(\therefore \cos 0^{°}=1\right)$

So, velocity of the vase = velocity at the end of free fall

$4.43={\omega }_n{x}_m$

Now, Natural frequency: ${\omega }_n=\sqrt{\frac{k}{m}}$

$\begin{array}{l}{\omega }_n=\sqrt{\frac{10000}{5}}\\ {\omega }_n=\sqrt{2000}\\ {\omega }_n=44.721rad/s\end{array}$

Thus, amplitude of the vase $x_m$ ,

$\begin{array}{l}4.43={\omega }_n{x}_m\\ x_m=\frac{4.43}{{\omega }_n}\\ x_m=\frac{4.43}{44.721}\\ x_m=0.09905mor,\\ x_m=99mm\end{array}$

And, maximum acceleration of the vase, $a={\omega }_n^2{x}_m$

$\begin{array}{l}a={\left(44.721\right)}^2\times 0.09905\\ a=198.096or,\\ a=198.096\times 0.10197g\\ a=20.2g\end{array}$

Conclusion:

Maximum acceleration of the vase is $a=20.2g$