# Vibration damping a bed from mechanical vibrations

Note: I understand the very nature of this question is about gross approximations on an incredibly complex topic, but even educated 'guesstimates' would be very helpful

I have already seen these two questions but i'm hoping with some provided data I can get some better insight.

I am experiencing constant mechanical vibrations in my bedroom from an unknown source somewhere in the building. They only last for 20-30mins then stop, but are an issue at night. It is the vibrations and not an actual noise causing issue. I am using vibration dampeners under my bed. I have tried 5 different products, with varying success. The current ones I am using seem to be 'styrene butadiene rubber' (medium hardness) with 'EVA polymetric foam' material' sandwiched between. Out of all the products so far, they are the softest and seem to have the best damping, although it is not enough.

However, I believe that all the products I am trying are still too hard for the frequency of the vibrations.

This idea is making me explore purchasing sorbothane disks (part numbers 0510435-30-10 or 0532350-50-10) with a 'duro' of either 30 (softer) or 50 (harder). I have found these spg and edg PDF files which show vibration damping calculations.

I am not an engineer and unfortunately I do not know the Hz of the vibrations I am experiencing. However, I can say it is akin to a washing machine near the end of a wash cycle on fast spin dry rather than a rumbling train.

The bed including myself and my partner weighs ~180 / ~200kg. The bed does not have four posts but a solid headboard and footboard which both reach the ground. This allows me to place more than 4 anti vibration pads rather than just one at each corner.

I am deciding between the two disks below (Imperial units listed);

0.25" Thickness x 2.25" Diameter (Duro 30, softer) 40lbs - 90lbs Load

6 Disks would give a max load of 244Kg (6 * (90 / 2.205))

0.50" Thickness x 2.25" Diameter (Duro 50, harder) 40lbs - 85lbs Load

6 Disks would give a max load of 231Kg (6 * (85 / 2.205))

So my main question; is a softer but thinner material better suited to absorbing high frequency vibrations than a thicker but slightly harder material?

From this answer, it implies that softer is better, but there is no mention of thickness.

And a bonus question;

Is it generally better to try 4 points of contact rather than 6? (thinner and harder, but two less points of contact) or should i actually try to increase the points of contact in order to use a softer material to dampen the vibrations?

EDIT: I have a response from one of the product development engineers from sorbothane who advises to go for a thicker part since thicker parts will have lower natural frequencies.

As to not waste the question, perhaps someone can answer why that is in either basic or technical terms.

• could you post an image of the current configuration? Additionally have you tried to stack up the different solution? Mar 29 at 14:06
• Posted an image
– myol
Mar 30 at 7:27
• I just replied. I hope it answers why, in basic and slightly technical terms (I thought that was the question). As you saw, in my previous comment I had asked if you tried stacking up the blocks. Mar 31 at 15:46

As you know you already had better results with softer compounds. So if you image the solid block as a spring, then you need a spring with a lower coefficient K. So you need to have higher displacement for a given force to follow hooke's law:

$$F = - K\cdot x$$

Now, the solid block isn't immediately recognizable as a spring, however you can estimate it as a spring with the following thought concept. If you apply a force F on a foam block then you will get a displacement.

Obviously, for the same displacement the force is greater:

• for greater Cross-section area A
• for greater material stiffness E
• for smaller thickness of the foam (lets call it length L).

From basic mechanics of material you get: $$K = \frac{EA}{L}$$

So you as the thickness increases, the spring stiffness lowers.

## Vibrations.

For a simple harmonic oscillator (like the image below),

you can show that the natural frequency is

$$\omega_n = \sqrt{\frac{k}{m}}$$

I.e. when you increase the K, you increase the natural frequency and vice versa. So what you are doing by increasing the thickness is reducing the natural frequency.

## Transmissibility coefficient

So why do you want to lower the natural frequency. In a simple DOF system (your bed is not really), the system gets "excited" the most at a frequency near the natural frequency. The following graph shows that with the help of the frequency ratio $$\Omega$$.

$$\Omega$$ is defined as $$\frac{\omega}{\omega_n}$$. So for $$\Omega=1$$, $$\omega=\omega_n$$.

And $$\omega$$ is the excitation frequency (i.e. the frequency of the vibrations under your bed). So as long as the frequency of the vibration is constant and near $$\omega_n$$ you'll get a headache (to put it simply).

Since you can't control the $$\omega$$ your best bet is to increase the frequency ratio and move to higher values of the frequency ratio $$\Omega$$. And the only way to do that is by lowering $$\omega_n$$

## 1. Main question: what is best

So my main question; is a softer but thinner material better suited to absorbing high frequency vibrations than a thicker but slightly harder material?

In general from what I've shown above (since you already observed you have better results with softer materials), you should strive for lower overall stiffness. That means:

• softer material (E: modulus of elasticity)
• L (greater thickness of the material). This is the easiest to play with.

That should decrease the natural frequency and then the Transmissability ratio will have lower values. If I haven't explained it before, lower values means of the transmissibility ratio ($$T$$)means that you have lower vibrations. If for example, $$T$$ is 0.5, then if the floor is vibrating with 1[mm] amplitude, then your bed would vibrate by 0.5[mm].

## 2. Bonus question: Position of pads

You could try to move the pads, and see if you get any results. I think it might be best if you actually put as many as possible, but then again that might be expensive.

So before you buy anything: try the following. I would move the 2 pads away from the corners. I would put each pad at about 20% from the end. So if you have a bed that is 1.60[m], I would put the pads about 30-35 cm away from the edge.

You might see some difference if you do that.

• Thank you for this. In the second equation of root k over m, is m in this case the mass of the object? and is k the spring coefficient K that can be used from the first equation?
– myol
Apr 2 at 7:51
• And in terms of contact points, am I correct in reasoning that as the number of contact points increases, the cross section A will increase, which in turn will increase the spring coefficient K?
– myol
Apr 2 at 7:53
• and in turn, from your comment below the second equation, more contact points will actually increase the natural frequency? i.e. reducing the anti vibration effectiveness?
– myol
Apr 2 at 7:55
• @myol, regarding m and k yes. Regarding the second point and third point, increasing the contact points is a more complex matter, because you can have twist. If you want we could continue this in a discussion, because there are a few subtle points that cannot be properly conveyed in a comment. Apr 2 at 8:15
• To give you an idea, when you have more that one contact points then you can have rotations and twisting instead of only ups and downs. Whether the rotaton happens depends on many things, the type/source of the vibrations, the distance and mounting. Additionally there are ways that you can have all 4 points behave as one (i.e. put a stiff massive block underneath the pads and the bed). In general though, this is the definition of "The devil is in the details". As I said before, I would start by stacking additional pads, it will get softer (see springs in series). Apr 2 at 14:24

First of all, before looking at any solution, you need to characterize two things

• the frequency of the vibrational input, which you can measure by placing an accelerometer on the floor. There is a quite good 3axis accelerometer built into every smart phone, so download the appropriate app (there are free ones that are more than good enough), take measurements, and do the FFT
• the vibrational response of the bed itself, with a mass (1 or perhaps 2 people). You can again observe that very easily with your cel phone, by doing a "step response" e.g. by sitting down on the bed.

I suspect that what you are feeling is in large part the resonance of your bed itself more so than the most significant part of the original excitation spectrum. But it could be a combination of the two. (spend a night sleeping on the floor on a camping mat to figure out which is the more significant one).

Finally, you can come up with a strategy to do something about it. Here is the big problem: to actually isolate the bed from the floor, the damping device has to have not only the damping ratio, but also a low enough spring constant, relative to the mass. If it has excellent damping ratio but too high spring constant/mass, it will simply behave like a solid object at the frequencies of interest and might as well be a rock. HOWEVER! if you give it a low spring constant/mass, there has to be vertical space for deflection when you sit on the bed etc.

Personally I think it is unlikely that an item like the one pictured would be useful for the situation you describe, because it is simply not tall enough to have room for the fairly low spring constant/mass that is needed. But start by making the measurements.

• Yes measurements would definitely help but it seems my phone is not sensitive enough. That said, I will try a combination of sleeping arrangements since that is in my control.
– myol
Apr 2 at 7:58