
In this lecture, we'll take a closer look at the steady-state response; recall the equation from the last lecture,

$$
u_p(t) = \frac{P_0}{k}\left[\frac{1}{(1-\beta^2)^2+(2\xi\beta)^2}\right]\left[(1-\beta^2)\sin(\omega t)-2\xi\beta\cos(\omega t)\right]
\tag{1}
$$

We notice that the response has two components, one in phase with the forcing function,

$$
\frac{P_0}{k}\left[\frac{1}{(1-\beta^2)^2+(2\xi\beta)^2}\right]\:\times \:(1-\beta^2)\sin(\omega t)
\tag{2}
$$

and one $90^\circ$, out of phase with it (indicated by the use of the cosine term),

$$
\frac{P_0}{k}\left[\frac{1}{(1-\beta^2)^2+(2\xi\beta)^2}\right]\:\times \:-2\xi\beta\cos(\omega t)
\tag{3}
$$

The out of phase component is introduced by the fact that there is damping in the system. We can think of this damping as (phase) shifting the response out of phase with the applied force causing the response to 'lag' behind the force. This lag is seen as a phase difference between applied force and response.

A helpful way to think about the steady-state response is in terms of its relationship to the forcing function that is generating it. To get started we can represent the two components of the steady-state response as rotating vectors on the complex plane, but before we sketch this out, we'll make things easier to read by making the following temporary substitution,

$$
\Omega = \frac{P_0}{k}\left[\frac{1}{(1-\beta^2)^2+(2\xi\beta)^2}\right]
\tag{4}
$$

Now, we can visualise both components of the steady-state response as two vectors, $90^\circ$ out of phase, rotating clockwise with amplitudes given as indicated in Fig. 1.

<Image
	src="/images/courses/fundamentals-of-structural-dynamics-with-python/characterising-the-steady-state-response/img1.jpg"
	alt="Two components of the steady-state response represented as rotating vectors on the complex plane. | EngineeringSkills.com"
	width="400"
	height="397"
	caption="Fig 1. Two components of the steady-state response represented as rotating vectors on the complex plane."
/>

The red vector $R$ represents the actual steady-state response and is the vector sum of the two (black) component vectors. The magnitude of the response at any instant in time can be obtained as the projection of the red vector onto the real axis.

Remember that the applied force is in phase with the leading response vector with magnitude $\Omega(1-\beta^2)$. So, the phase angle $\phi$ represents the degree to which the system response (red vector) lags behind the applied harmonic force.

Also notice that as the level of damping increases, the length of the trailing vector (with magnitude $\Omega2\xi\beta$) increases and therefore the resultant response vector will lag further behind the applied force.

The resulting vector $R$ has magnitude,

$$
|u(t)|=\Omega \sqrt{(1-\beta^2)^2 + (2\xi\beta)^2}
\tag{5}
$$

Reversing the temporary substitution (for $\Omega$) we just made yields,

$$
|u(t)|=\frac{P_0}{k}\frac{1}{\sqrt{(1-\beta^2)^2 + (2\xi\beta)^2}}
\tag{6}
$$

Note that $P_0/k$ is the equivalent static deflection and the term to the right of it in equation 6 is essentially a magnification factor. This is something we'll continue to discuss in the next lecture.

We can also state the phase of the steady-state response with reference to the component of response in phase with the force as,

$$
\boxed{
\phi = \tan^{-1}\frac{2\xi\beta}{1-\beta^2}
}
\tag{7}
$$

We will continue to explore the phase and amplitude of the steady-state response, in comparison to the forcing function, in the next lecture.


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