Chapter 11, Problem 45P

To determine

To Find: The amplitude of the object.

Answer to Problem 45P

  $0.99\text{ m}$

Explanation of Solution

Given info:

Frequency $f=0.5\text{ Hz}$

Formula used:

The expression for acceleration.

  $a_{\max }=A\times {\omega }^2$

  $a_{\max }$ is greater than the acceleration due to gravity.

Thus,

  $\begin{array}{l}A\times {\omega }^2>g\\ A>\frac{g}{{\omega }^2}\end{array}$

Here,

  $g$ is acceleration due to gravity.

  $A$ is amplitude of oscillation.

  $\omega$ is angular frequency.

Calculation:

Assume the earthquake wave moving the ground vertically since it is a transverse wave. Object sitting on the ground would be moving with some harmonic motion. The object has normal force and self weight.

If object losses contact with the ground the normal force of an object is equal to zero. Only force acting on the object is self weight. Substitute the given values:

  $\begin{array}{l}A>\frac{g}{{\omega }^2}=\frac{g}{4\times {\pi }^2\times f^2}\\ \text{As},\text{}\omega =2\pi f\end{array}$

Limiting factor of amplitude, $A=\frac{9.8}{4\times {\pi }^2\times {0.5}^2}\text{= }0.99\text{ m}$

Conclusion:

Thus, the amplitude in such a way that the object begins to leave contact with the ground is $0.99\text{ m}$ .