Now, let's look at again this mathematical expression in

terms of transfer function.

Let x(t) has complex amplitude

and exponential j omega t.

Also, F of t has complex amplitude exponential j omega.

Then, this equation will be expressed like

minus omega square mx plus,

interestingly j omega c.

Because I am differentiating this once,

so I have j omega coming out,

and the c exponential j omega t,

plus kx equal to F.

Because we have the

same exponential j omega t, we discarded it.

Now, this mathematical expression shows that, okay,

there is relation between

complex amplitude and F.

That is one over minus omega square m

plus j omega c plus kx.

If you take the magnitude of this,

you will get the transfer function magnitude.

If you see the face of this,

we will see the transfer function phase

of the transfer function as we saw before.