Chapter 9.5, Problem 41E

To determine

To find:Relationship between the constants $C_1\&C_2$

Answer to Problem 41E

The relation is $C_1=C_2$ .

Explanation of Solution

Given information :

Equation of motion in spring mass system $C_1(\cos 2t+i\sin 2t)+C_2(\cos 2t-i\sin 2t)$

Concept used:

Equation of motion using spring mass system involves solution of differential equation in form of $\cos \beta t+i\sin \beta t$

Equation of motion in spring mass system $C_1(\cos 2t+i\sin 2t)+C_2(\cos 2t-i\sin 2t)$ , for all values of $t$ .

  $\begin{array}{l}{C}_1(\cos 2t+i\sin 2t)+C_2(\cos 2t-i\sin 2t)\\ \Rightarrow C_1\cos 2t+C_{1}i\sin 2t+C_2\cos 2t-C_{2}i\sin 2t)\\ \Rightarrow (C_1+C_2)\cos 2t+(C_1-C_2)(i\sin 2t)\end{array}$

We have $\mathrm{Re}+i\mathrm{Im}$

This is real only if the imaginary component is zero,

And this happens when $C_1=C_2$

Thus, sum equals to $(C_1+C_2)\cos 2t$

Hence,

The relation is $C_1=C_2$ .