# Mass atop four springs with external input: Reduce vibration amplitude by factor of ten

• zenterix
zenterix
Homework Statement
Suppose we have a large delicate material of mass ##M## supported by four massless springs in parallel, each with spring constant ##k##.

The whole setup is put on a table.

When the table is moved across the floor, the tabletop vibrates, producing an effective vertical force

$$F=MA_0\cos{\omega d t}$$

on the mass ##M## in the tabletop reference frame.
Relevant Equations
a) Write down the equation of motion of the instrument. Assume no drag force in this part of the question.

b) Find the vibration amplitude of the mass in the steady state.

c) What would you do to reduce the vibration amplitude by a factor of ten?

Assume ##k/M\gg\omega_d^2##. Hint: the spring constant is proportional to its area and inversely proportional to its length.

d) A better way to reduce the vibration amplitude is to introduce damping: insert some kind of massless cushion between the mass and the table in parallel to the springs. Assuming the cushion produces a resistive force ##-b## times the velocity of ##M##, derive an equation that allows you to determine the value of ##b## in terms of ##k,M,\omega_d## and solve for ##b##, for ##k/M\gg\omega_d^2##.
This is problem 2 from problem set 2 of MIT OCW's 8.03 "Vibrations and Waves". You can see the pset here. There are no solutions available from MIT OCW.

The force on the mass is ##F_s=-4ky##.

Thus

$$M\ddot{y}=-4ky+F_{table}$$

$$\ddot{y}+\omega_0^2y=A_0\cos{\omega_d t}$$

where ##\omega_0^2=\frac{4k}{M}##

A solution to this differential equation is

$$y(t)=\frac{A_0}{\omega_0^2-\omega_d^2}\cos{\omega_d t}$$

and we see that the vibration amplitude is ##\frac{A_0}{\omega_0^2-\omega_d^2}##.

To reduce this by a factor of ten we need

$$\frac{A_0}{\omega_1^2-\omega_d^2}=\frac{1}{10}\frac{A_0}{\omega_0^2-\omega_d^2}$$

where ##\omega_1=\frac{4k_1}{M}##. That is, it is the natural frequency of the SHM of the mass when we have springs with a new constant ##k_1##.

We can write ##k=C\frac{a}{l}##, ie ##k## is proportional to area ##a## and inversely proportional to length ##l## with proportionality constant ##C##.

Suppose the new springs scale the area of the initial springs by ##v## and the length by ##u##. Then ##k_1=C\frac{va}{ul}##.

Subbing in we reach the expression

$$\frac{v}{u}=10-\frac{9l}{4a}\frac{1}{C}M\omega_d^2$$

If we pick ##v## and ##u## satisfying this equation then we get the desired reduction in the vibration amplitude.

The equation of motion becomes

$$\ddot{y}+\frac{b}{M}\dot{y}+\frac{4k}{M}y=A_0\cos{\omega_d t}$$

$$\ddot{y}+\gamma\dot{y}+\omega_0^2y=A_0\cos{\omega_d t}$$

which has a solution

$$y(t)=\frac{A_0}{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega_d^2}\cos{(\omega_d t-\phi)}$$

where

$$\tan{\phi}=\frac{\gamma\omega_d}{\omega_0^2-\omega_d^2}$$

The vibration amplitude is ##\frac{A_0}{(\omega_0^2-\omega_d^2)+\gamma^2\omega_d^2}## which is smaller than before.

For this amplitude to be ##1/r## of the previous amplitude we need

$$\frac{A_0}{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega_d^2}=\frac{1}{r}\frac{A_0}{\omega_0^2-\omega_d^2}$$

$$b=M\sqrt{r\left ( \frac{\omega_0^2}{\omega_d^2}-1\right )-\left ( \frac{\omega_0^2}{\omega_d^4}-\frac{1}{\omega_d^2} \right )^2}$$

We are interested in the case of ##r=10##.

My question is about the assumption that ##\frac{k}{M}\gg\omega_d^2##.

Why do we need this assumption?

In addition, it is not clear that the expression I found for ##b## is correct when I take into account this assumption. It seems what is inside the square root can be negative.

For example, suppose

$$M=1\text{kg}$$

$$r=10$$

$$\omega_d=1$$

Then a plot of ##b(\omega_0)=\sqrt{(\omega_0^2-1)(11-\omega_0^2)}## is

Last edited:
zenterix said:
Why do we need this assumption?

Hi,

Your $$\frac{A_0}{\omega_1^2-\omega_d^2}=\frac{1}{10}\frac{A_0}{\omega_0^2-\omega_d^2}$$ then simplifies to ##k_1\approx 10 k##.

##\ ##

Steve4Physics
Just to expand on what @BvU said in Post#2 (though it may be unnecessary)…

Using ##\frac kM \gg \omega_d^2## reduces the amount of work you need to do, since it allows approximations to be made.

Since ##\omega_0^2 = \frac {4k}M## and ##\frac kM \gg \omega_d^2##, this means ##\omega_0^2 \gg \omega_d^2##.

Then, for example in part c), the amplitude is ##\frac{A_0}{\omega_0^2-\omega_d^2}## so this can be approximated to ##\frac{A_0}{\omega_0^2}##. It then immediately follows that to reduce the amplitude by a factor of 10 we need to increase ##k## by a factor of 10 (as already stated by @BvU).

(But sorry if that was already understood!)

Steve4Physics said:
It then immediately follows that to reduce the amplitude by a factor of 10 we need to increase k by a factor of 10
It seems that multiplying ##k## by 10 accomplishes the result always when ##\omega_0>\omega_d##, which makes the denominators in the first expression below positive.

$$\frac{A_0}{10\omega_0^2-\omega_d^2}<\frac{A_0}{10\omega_0^2-10\omega_d^2}$$

$$\implies 10\omega_0^2-\omega_d^2>10\omega_0^2-10\omega_d^2$$

$$\implies \omega_d^2<10\omega_d^2$$

$$\implies 1<10$$

Here is a visual of this

In the plots above, I assume ##y=\omega_d=1##. For any ##\omega_0>1## we see that the red line is below the blue line (this isn't totally clear from the plot but should be from the equations barring a mistake).

I think it may make sense to look at the absolute values

Note that the blue curve represents, for each ##\omega_0##, one tenth the amplitude of the initial vibration for that ##\omega_0##.

The red curve represents, for each ##\omega_0##, the amplitude when we multiply ##k## by 10 (ie, when we multiply ##\omega_0^2## by 10).

Near resonance, the original amplitude and one tenth of it becomes very large.

The resonance of the spring with a ten times higher ##k## happens at a much lower natural frequency.

We need to get away from this natural frequency for the amplitude to be below ##1/10## the original amplitude.

In fact, the point where the two curves cross is at ##\omega_0\approx 0.74##.

In general (ie, when we don't assume, as above, that ##\omega_d=1##) the point where the absolute values of the amplitudes cross is

$$0.74\omega_d,\ \ \ \ \ \ \text{assuming}\ \omega_d>0$$

For clarity, this came from

That is, given a driving frequency and a natural frequency (ie given an initial spring), multiplying the spring constant by 10 reduces the amplitude below 1/10 the original amplitude if ##\omega_0>0.74\omega_d##.

If this condition is not met, then we would need to multiply ##k## by a larger number.

As a final point, what my original calculation did, in terms of the plots above, was the following.

Given ##\omega_0## and ##\omega_d##, draw a vertical line in the plot above at ##\omega_d##. Then, what is the exact factor to multiply ##k## by so that when we plot ##\frac{A_0}{k\omega_0^2-\omega_d^2}## the resulting (red) curve crosses the point where the vertical line crosses the plot of ##\frac{1}{10}\frac{A_0}{\omega_0^2-\omega_d^2}##, ie the blue curve.

Last edited:

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