Chapter 11, Problem 11P

To determine

To Find: The distance at which the acceleration is half of its maximum acceleration.

Answer to Problem 11P

The distance from the equilibrium where acceleration is half of its maximum acceleration is $0.5x_0$ .

Explanation of Solution

Given info:

  $\text{Acceleration }a=\frac{1}{2}\times a_{\max }$ .

Formula used:

Expression of force :

  $F=-k\times x=m\times a$

Here,

  $F$ is force,

  $x$ is the displacement of spring,

  $k$ is spring constant,

  $m$ is mass on spring,

  $a$ is acceleration of spring.

Calculation:

Substitute the given values in the above equation:

  $\begin{array}{l}F=-k\times x=m\times a\\ a=-\frac{k\times x}{m}\end{array}$

As, it is clearly seen from the above equation, acceleration is directly proportional to distance, so if acceleration is equal to $0.5a_{\max }$ only when the distance is half the available distance that is $x=0.5x_0$ .

Conclusion:

Thus, the distance from the equilibrium where the acceleration is half of its maximum acceleration is $0.5x_0$ .