Amplitude of the oscillation

A tank contains 201 gallons of water and 49 oz of salt. Water containing a salt concentration of $\displaystyle \frac{1}{7}\left(1+\frac{1}{3}sin(t)\right)$ oz per gal flows into the tank at a rate of 2 gal per min, and the mixture in the tank flows out at the same rate.The long-time behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation?

Could someone help me out with this question? My textbook has this as a problem, but with no example in the chapter to follow.

Re: Amplitude of the oscillation

I would expect an example matching an exercise in a text to be in the preceding chapter, not the "chapter to follow"!

Of course, you don't even say what type course this is so I have no idea what methods you might have available. But here is how **I** would do it:

Let X(t) be the total amount of salt. in oz, in the tank at time t, on minutes. Then $\displaystyle \frac{dX}{dt}$ is the rate at which salt is coming into the the tank. We are told "$\displaystyle \frac{1}{7}\left(1+ \frac{1}{3}sin(t)\right)$ oz of salt per gallon come into the tank at 2 gallons per minute" so $\displaystyle \frac{2}{7}\left(1+ \frac{1}{3}sin(t)\right)$ oz of salt come into the tank every minute. With X oz of salt in a 201 gal tank, there are $\displaystyle \frac{X}{201}$ oz of salt in each gallon so if the mixture goes out at 2 gallons per minute, there are $\displaystyle \frac{2X}{201}$ oz of salt leaving every minute.

That is, we have the equation $\displaystyle \frac{dX}{dt}= \frac{2}{7}\left(1+ \frac{1}{3}sin(t)\right)- \frac{2X}{201}$. That is equivalent to the first order linear differential equation $\displaystyle \frac{dX}{dt}+ \frac{2}{201}X= \frac{2}{7}\left(1+ \frac{1}{3}sin(t)\right)$. (Is this a differential equations class?) The associated "homogeneous equation" is $\displaystyle \frac{dX}{dt}+ \frac{2}{201}X= 0$ and its general solution is $\displaystyle Ce^{-\frac{2}{201}t$. That obviously will go to 0 very quickly and is called the "transient solution" by engineers. To find the "particular solution", the one that "lasts" (and is called the "steady state solution") look for a solution of the form $\displaystyle A+ Bsin(t)+ Ccos(t)$. The "$\displaystyle Bsin(t)+ Ccos(t)$" is the "oscillation" of course, and is not relevant to the first part of this exercise which asks for the "constant level" the solution oscillates about and that is the number "A".

So- if you let X(t)= A, a constant, what value of A satisfies the differential equation? Obviously, $\displaystyle \frac{d^2A}{dt^2}= 0$ and the "$\displaystyle \frac{1}{3}sin(t)$" is the "$\displaystyle Bsin(t)+ Ccos(t)$" part of the equation. Finding the "constant level the solution oscillates about" reduces to solving $\displaystyle \frac{2}{201}A= \frac{2}{7}$!

To find the amplitude of the oscillation means finding the amplitude of the function Bsin(t)+ Ccos(t) and that, of course, depends on B and C. Substitute X= B sin(t)+ C cos(t) into $\displaystyle \frac{dX}{dt}+ \frac{2}{201}X= \frac{2}{21}sin(t)$ to determine what B and C must be.

Re: Amplitude of the oscillation

Quote:

Originally Posted by

**HallsofIvy** I would expect an example matching an exercise in a text to be in the preceding chapter, not the "chapter to follow"!

That's what I meant. The way I stated it is a little ambiguous, so my apologies.

Thanks for the explanation though. It was extremely helpful.

Re: Amplitude of the oscillation

$\displaystyle X=\frac{201}{7} + \frac{\frac{2}{21}\left(\frac{2}{201}sin(t)-cos(t)\right)}{\left(\frac{4}{201^2}+1\right)^{1/2}}+ce^{\frac{-2}{201}t}$