Determine the Maximum Speed of an Oscillating System: A Bumpy Road

We can use the expression for $v_{\text{max}}$ given in $v_{\text{max}}=\sqrt{\frac{k}{m}}X$ to determine the maximum vertical velocity. The variables $m$ and $k$ are given in the problem statement, and the maximum displacement $X$ is 0.100 m.

1. Identify known.
2. Substitute known values into $v_{\text{max}}=\sqrt{\frac{k}{m}}X$:
   $$v_{\text{max}}=\sqrt{\frac{6\text{.}\text{53}\times {\text{10}}^4\text{N/m}}{\text{900}\text{kg}}}\mathrm{(0}\text{.}\text{100}\text{m)}.$$

   16.44
3. Calculate to find $v_{\text{max}}\text{= 0.852 m/s}.$

This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find $v_{\text{max}}$. We could use it directly, as was done in the example featured in Hooke’s Law: Stress and Strain Revisited.

The small vertical displacement $y$ of an oscillating simple pendulum, starting from its equilibrium position, is given as

where $a$ is the amplitude, $\omega$ is the angular velocity and $t$ is the time taken. Substituting $\omega =\frac{\mathrm{2\pi }}{T}$, we have

Thus, the displacement of pendulum is a function of time as shown above.

Also the velocity of the pendulum is given by

so the motion of the pendulum is a function of time.