Difference between these frequency response curves

Question

I am thinking about a simple mass-damper system response. Most single DOF systems show a frequency response plot that looks like this:
Note how increasing the damping ratio always reduces the output response, and at high frequencies, the response approaches the same line regardless of damping ratio.

But, the way I am used to thinking about it, the response should really look like this:

Note how the increasing the damping ratio reduces the response below $\omega/\omega_0 \cdot \sqrt{2}$, but increases at frequencies above $\omega/\omega_0 \cdot \sqrt{2}$.
The response does not approach a downward sloping line as damping ratio increases. Rather, it approaches $1$ as damping ratio increases.

So what am I missing here? What is the difference in these two curves?

Thanks

Answer 1

The two frequency response plots are the result of two different models. In the first one, the model is a mass-spring-damper system as shown here, and the plot shows the ratio of the amplitudes of mass displacement and driving force.

The second plot shows the transmissibility of a single DOF mechanical system as shown below:

(from Fundamentals of Mechanical Vibrations by S.G.Kelly).

Note that here, unlike in the other model, the support is not fixed. The transmissibility shown in your second plot is the ratio of the amplitudes of the mass displacement $x(t)$ and the (forced) displacement of the support $y(t)$.

In formulas, your first plot shows the function

$$M(\omega)=\frac{1}{\sqrt{\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)^2+\left(2\zeta\frac{\omega}{\omega_0}\right)^2}}\tag{1}$$

where $\omega_0$ is the system's natural frequency, and $\zeta$ is the damping ratio. In electrical engineering terms, this is the magnitude of the frequency response of a second-order low pass filter.

The second plot shows the transmissibility given by Eq. (3.73) in Kelly's book:

$$T(\omega)=\sqrt{\frac{1+\left(2\zeta\frac{\omega}{\omega_0}\right)^2}{{\left(1-\left(\frac{\omega}{\omega_0}\right)^2\right)^2+\left(2\zeta\frac{\omega}{\omega_0}\right)^2}}}\tag{2}$$

Note that $(1)$ and $(2)$ have the same denominator but different numerators.