In a problem I came across I was asked to determine the parameters M2 and k12 such that mass M1 does not vibrate in steady state. The diagram can be seen below

I found the transfer function to be
Y1(s) = aM2w(s2 + k12 / M2) / ((s2 + w2) * delta) where delta is the determinant of the matrix.

This transfer function is definitely correct, but I tried for ages and ages to determine M2 and k12, whilst setting Y1(s) to 0, to no success. The answer booklet gets
w2 = k12/M2
and does this by setting
(s2 + k12 / M2) = (s2 + w2). I've attached a picture of this below.

I dont know how or why they set these equal to each other.

• Please edit the question to include a diagram in case the original problem came with a diagram. It is not clear what $M_1$ is or what $w$, $\textrm{DELTA}$ are or what are the inputs and outputs of this system or what $Y_1$ physically is. In case there is a blck diagram, please include that too. Please also describe what physics the transfer function represents.
– AJN
Apr 2, 2023 at 3:06
• Thank you for your respond sorry for being unclear about the question. Hopefully, it should be clear from the diagrams what M1 and delta is.
– Anna
Apr 2, 2023 at 9:08
• I'm not sure I know what you mean by the LHS of the s-plane? I'm not familiar with the term. Thank for your help again
– Anna
Apr 2, 2023 at 18:18
• Ignore my previous comment.
– AJN
Apr 2, 2023 at 18:46

The system appears to be a tuned mass damper. Image from Wikipedia.

If we look at the response $$Y_1$$, we see that the denominator has two parts viz; $$(s^2 + \omega_0^2)$$ and $$\Delta(s)$$. The masses, springs and dampers constitute a passive system and so the steady state response contribution from $$\Delta(s)$$ is zero. However, the steady state contribution from $$(s^2 + \omega_0^2)$$ is not zero (example). By tuning the numerator to have the same term, $$(s^2 + \omega_0^2)$$ in the denominator gets cancelled and the response becomes dependent only on $$\Delta(s)$$. This allows the steady state value of $$M_1$$'s motion to become to become zero.

• So in steady state you have to make the contributing factors i.e. 𝑠2+𝜔20 and s^2+k_12/M_2 equal to each other to insure they cancel?
– Anna
Apr 2, 2023 at 19:18
• Yes. The factor in the denominator is the problematic term that needs to be cancelled out. The factor in the numerator which has the same format is used for this purpose. Are you familiar with partial fraction expansion, inverse Laplace transform and final value theorem? Those are the tools which may be useful in understanding the above.
– AJN
Apr 2, 2023 at 19:20
• Yes, we have been taught this but I dont see how it relates to this case. So instead of setting the transfer function equal to zero we can just identify the contributing factors to find the steady state solutions?
– Anna
Apr 2, 2023 at 20:12
• You are effectively setting the transfer function to zero; but only at the frequency $\omega_0$.
– AJN
Apr 3, 2023 at 13:05