Question

Compare the result obtained in part (b) of Problem 43 with the solution obtained using variation of parameters when the external force is ${\mathit{F}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}$.

Short answer

On comparing the result is $x\left(t\right)=\frac{F_0}{2\omega }t\sin \gamma t$.

Step-by-step solution

Definition of external force:

External forces are forces resulting from the interaction between human body and its environment. External forces can be divided into contact forces and non-contact forces. Most forces that biomechanics works with are contact forces.

Substituting the given differential equation and comparing the coefficient:

As you clearly know that,

$$\begin{gathered} A=0 \\ B=\frac{F_0}{2\omega } \\ {x}_p\left(t\right)=\frac{F_0}{2\omega }t\sin \gamma t \end{gathered}$$

The general solution is,

$x\left(t\right)=c_1\cos \omega t+c_2\sin \omega t+\frac{F_0}{2\omega }t\sin \gamma t$

Using the initial condition to solve the equation,

$x\left(0\right)=0,x\text{'}\left(0\right)=0$

And you get,

$$\begin{gathered} c_1=c_2=0 \\ x\left(t\right)=\frac{F_0}{2\omega }t\sin \gamma t \end{gathered}$$

Hence, the final answer is $x\left(t\right)=\frac{F_0}{2\omega }t\sin \gamma t$.