- #1

zenterix

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- 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.

Next, consider adding damping.

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

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

which leads to

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

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

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.

Next, consider adding damping.

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

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

which leads to

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

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

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