A machine of mass 100 kg is subjected to an external harmonic force with a frequency of 40 rad/s. The designer decides to mount the machine on an isolator to reduce the force transmitted to the foundation. The isolator can be considered as a combination of stiffness (K) and damper (damping factor, ζ) in parallel. The designer has the following four isolators :

1) k = 640 kN/m, ζ = 0.70

2) k = 640 kN/m, ζ = 0.07

3) k = 22.5 kN/m, ζ = 0.70

4) k = 22.5 kN/m, ζ = 0.07

Arrange the isolators in the ascending order of the force transmitted to the foundation.

The correct option is "4-3-1-2".

1. Understanding Transmissibility and the System Parameters
The force transmitted to the foundation depends on the transmissibility ratio ($\epsilon$), defined as the ratio of the transmitted force ($F_t$) to the excitation force ($F_0$). The transmissibility ratio is given by:

$$\epsilon =\frac{F_t}{F_0}=\sqrt{\frac{1+{\left(2\zeta r\right)}^2}{{\left(1-r^2\right)}^2+{\left(2\zeta r\right)}^2}}$$

where:
• $\zeta$ is the damping ratio (damping factor),
• $r$ is the frequency ratio, defined as $r=\frac{\omega }{{\omega }_n}$, with excitation frequency $\omega =40\text{ rad/s}$ and natural frequency ${\omega }_n=\sqrt{\frac{k}{m}}$ (for a mass $m=100\text{ kg}$).

2. Calculating the Frequency Ratio ($r$) for the Two Stiffness Values
From the analysis in the provided diagram, we compute the parameters for the two stiffness settings:

Case A: For $k=640\text{ kN/m}=640,000\text{ N/m}$ (Isolators 1 and 2):

$${\omega }_n=\sqrt{\frac{640,000}{100}}=80\text{ rad/s}$$

$$r=\frac{40}{80}=0.5$$

Since $r=0.5<\sqrt{2}$, these isolators operate in the region below the crossover point.

Case B: For $k=22.5\text{ kN/m}=22,500\text{ N/m}$ (Isolators 3 and 4):

$${\omega }_n=\sqrt{\frac{22,500}{100}}=15\text{ rad/s}$$

$$r=\frac{40}{15}\approx 2.667$$

Since $r\approx 2.667>\sqrt{2}$, these isolators operate in the isolation region above the crossover point.

3. Analyzing Transmissibility Curves from the Diagram
As shown in the plotted graph of transmissibility ($\epsilon$) versus frequency ratio ($r$):
• For $r<\sqrt{2}$: The transmissibility is greater than 1 ($\epsilon >1$). In this region, increasing the damping factor ($\zeta$) decreases the transmissibility. Comparing Isolator 1 ($\zeta =0.70$) and Isolator 2 ($\zeta =0.07$):

$${\epsilon }_1<{\epsilon }_2$$

• For $r>\sqrt{2}$: The transmissibility is less than 1 ($\epsilon <1$). In this isolation region, increasing the damping factor ($\zeta$) actually increases the transmissibility. Comparing Isolator 3 ($\zeta =0.70$) and Isolator 4 ($\zeta =0.07$):

$${\epsilon }_4<{\epsilon }_3$$

4. Establishing the Ascending Order
Since the transmissibility values for the isolation region ($r>\sqrt{2}$) are less than 1, and the values for the region ($r<\sqrt{2}$) are greater than 1, we can order the transmissibility values directly:

$${\epsilon }_4<{\epsilon }_3<1<{\epsilon }_1<{\epsilon }_2$$

Thus, the ascending order of the force transmitted to the foundation is:
4 - 3 - 1 - 2