Chapter 11, Problem 87GP

To determine

To find: The period of bouncing.

Answer to Problem 87GP

The period of bouncing is $0.40\text{s}$ .

Explanation of Solution

Given:

Mass of the car $m=1200\text{kg}$

  $\begin{array}{l}L=22\text{m}\\ d=(6.4\times {10}^{-3})\text{m}\end{array}$

Concept used:

According to Hooke’s law

  $\begin{array}{l}E=\frac{\text{stress}}{\text{strain}}\\ E=\frac{F/A}{\Delta L/L}\end{array}$

By rearranging,

  $F=\frac{EA}{L}\times \Delta L$

  $K=\frac{EA}{L}$

The time period is given as,

  $\begin{array}{l}T=2\pi \sqrt{\frac{m}{K}}\\ T=2\pi \sqrt{\frac{mL}{EA}}\end{array}$

By putting all the known values from the above,

  $\begin{array}{l}T=2\pi \sqrt{\frac{(1200\text{kg})(22\text{m})}{(200\times {10}^9{\text{N/m}}^{\text{2}})(\pi )(3.2\times {10}^{-3}\text{m})}}\\ T=0.40\text{s}\end{array}$

Conclusion:

The period of bouncing is $0.40\sec$ .