Chapter A6, Problem 68E

(a)

To determine

To show: that the acceleration is proportional to $s$

Answer to Problem 68E

The acceleration is proportional to $s$ . The negative constant of proportionality implies that the acceleration is directed towards the origin.

Explanation of Solution

Given:

  $s=a\cos kt+b\sin kt$

Calculation:

Given with the equation for displacement as

  $\begin{array}{l}s\left(t\right)=a\cos kt+b\sin kt\\ \Rightarrow \frac{ds}{dt}=-ak\sin kt+bk\cos kt\\ \Rightarrow \frac{d^{2}s}{d{t}^2}=-a{k}^2\cos kt-b{k}^2\sin kt\\ \Rightarrow \frac{d^{2}s}{d{t}^2}=-k^2\left(a\cos kt+b\sin kt\right)\\ \Rightarrow \frac{d^{2}s}{d{t}^2}=-k^{2}s\left(t\right)\end{array}$

This shows that acceleration is proportional to $s$ . the negative constant of proportionality implies that the acceleration is directed towards the origin.

Conclusion:

Therefore, the acceleration is proportional to $s$ .

(b)

To determine

To show: that the acceleration is proportional to $s$

Answer to Problem 68E

The acceleration is proportional to $s$ . The positive constant of proportionality implies that the acceleration is directed away from the origin.

Explanation of Solution

Given:

  $s=a\cosh kt+b\sinh kt$

Calculation:

Given with the equation for displacement as

  $\begin{array}{l}s\left(t\right)=a\cosh kt+b\sinh kt\\ \Rightarrow \frac{ds}{dt}=-ak\sinh kt+bk\cosh kt\\ \Rightarrow \frac{d^{2}s}{d{t}^2}=-a{k}^2\cosh kt-b{k}^2\sinh kt\\ \Rightarrow \frac{d^{2}s}{d{t}^2}=-k^2\left(a\cosh kt+b\sinh kt\right)\\ \Rightarrow \frac{d^{2}s}{d{t}^2}=k^{2}s\left(t\right)\end{array}$

This shows that acceleration is proportional to $s$ . The positive constant of proportionality implies that the acceleration is directed away from the origin.

Conclusion:

Therefore, the acceleration is proportional to $s$ .