Motion of a particle performing damped vibrations

[stu]

Hi

I have two questions to ask and hopefully someone could help as I am getting little help from my college and work collegues

1) The work done by an air compressor is given by
W = K [ (p1/p2)^(n-1/n) + (p1/p2)^(n-1/n)-2]

where p1,p2,n and K are all constants.
QUESTION - show that the work done is a minimum when p=sqrt(p1p2)

I have no idea where to start on this at all



2) motion of a particle performing damped vibrations is given by x=e^-t * sin 2t (x=displacement in metres and t=time in secs)
Find the times at which x is at a maximum and find the maximum distance for the least positive value of t. Find the acceleration at this point

I have found dx/dt and evaluated it for 0 This gives me an answer of t=31.6 (approx) but when i plug t=31.6 into dx/dt i come out with it equals 0, which suggests to me there is no maximum point but instead a point of inflection. That is about as far as i got on this question before i got stuck.
Any help on either of these question would be much appreciated as i have been trying to get my head around them for a while now with little or no success

Regards
Stuart

 

[Gokul43201]

1) There's something missing here...specifically there's no 'p' in the expression for the work.

 

[AKG]

Hi

I have two questions to ask and hopefully someone could help as I am getting little help from my college and work collegues

1) The work done by an air compressor is given by
W = K [ (p1/p2)^(n-1/n) + (p1/p2)^(n-1/n)-2]

where p1,p2,n and K are all constants.
QUESTION - show that the work done is a minimum when p=sqrt(p1p2)

I have no idea where to start on this at all

What's "p"? Do you mean n = \sqrt{p_1p_2}? Also, where exactly is that "-2"? Is it like this:

W = K \left [ { \left ( \frac{p_1}{p_2} \right ) }^{\left ( \frac{n-1}{n} \right ) } + { \left ( \frac{p_1}{p_2} \right ) }^{\left ( \frac{n-1}{n} \right ) } - 2 \right ]
W = 2K \left [ { \left ( \frac{p_1}{p_2} \right ) }^{\left ( \frac{n-1}{n} \right ) } - 1 \right ]
\frac{dW}{dn} = 2K \left [ \ln \left( \frac{p_1}{p_2} \right ){ \left ( \frac{p_1}{p_2} \right ) }^{\left ( \frac{n-1}{n} \right ) } (n^{-2}) \right ]

2) motion of a particle performing damped vibrations is given by x=e^-t * sin 2t (x=displacement in metres and t=time in secs)
Find the times at which x is at a maximum and find the maximum distance for the least positive value of t. Find the acceleration at this point

I have found dx/dt and evaluated it for 0 This gives me an answer of t=31.6 (approx) but when i plug t=31.6 into dx/dt i come out with it equals 0, which suggests to me there is no maximum point but instead a point of inflection. That is about as far as i got on this question before i got stuck.
Any help on either of these question would be much appreciated as i have been trying to get my head around them for a while now with little or no success

Regards
Stuart

There will be no points of inflection. If you graph this, you will see a sine wave with a decaying amplitude (that's what the e^{-t} does), hence "damped oscillation."

x = e^{-t} \sin (2t)

\frac{dx}{dt} = -e^{-t} \sin (2t) + 2e^{-t} \cos (2t)

0 = e^{-t} [2 \cos (2t) - \sin (2t) ]

0 = 2 \cos (2t) - \sin (2t)[\tex]<br /> <br /> 2 \cos (2t) = \sin (2t)<br /> <br /> \tan (2t) = 2<br /> <br /> 2t \approx 1.1071487<br /> <br /> t \approx 0.55357436<br /> <br /> <br /> x \approx 0.51419838

 

[Gokul43201]

2) Understand what the oscillations look like - it's a sine wave that whose amplitude diminishes exponentially with time.

a) Times at which x is at a maximum : This suggests that you are looking for local maxima, which can be found where dx/dt=0 or tan(2t) = 2.

You get the solutions 2t=tan^{-1}(2) + n\pi

b) This seems to be the first local maximum, which will be greater than all others because of the exponential decay factor. It happens at n=0 or t=Pi*31.6/180 (since you must use radians NOT degrees).

c) The acceleration is found by substituting this value of t into \frac{d^2x} {dt^2}

I have found dx/dt and evaluated it for 0 This gives me an answer of t=31.6 (approx) but when i plug t=31.6 into dx/dt i come out with it equals 0

Of course it would, wouldn't it ? Only if you get zero in the second derivative, is it a point of inflection.

 

[Gokul43201]

I think the -2 belongs inside the bracket. So the second exponent is (n-1)/(n-2)...only my guess ! Still...no p anywhere !

 

[stu]

calculus

hi folks, thanks for getting back so fast.

Firstly, the format of the question is exactly how AKG showed it (how did you do that btw?). Secondly i am aware that there is no P anywhere in the question, but that is exactly how it is worded and it is not n=sqrt(p1p2).

As for the second question i must of been watching the footy last night to much as i have rechecked my figures today and have now found a maxima point, but i am still not sure what "and find the maximum distance for the least positive value of t" means or is that just a long winded version of saying what the maxima point is?

Thanks again for the help

Stuart

 

[AKG]

hi folks, thanks for getting back so fast.

Firstly, the format of the question is exactly how AKG showed it (how did you do that btw?).

Search the forums for "LaTeX." There's a thread all about how to use it. Here is a pdf file that helps for starters.

Secondly i am aware that there is no P anywhere in the question, but that is exactly how it is worded and it is not n=sqrt(p1p2).

Well that's ridiculous! If it's for homework, I wouldn't bother with the question, it looks like a typo. What's more, even if it were n = \sqrt{p_1p_2} I don't think that works. The work I've shown so far on that problem leads to a result where no value of n will give a derivative of zero. Looking at the function itself, I don't see n = \sqrt{p_1p_2} being of any significance. It depends on the signs and magnitudes of the various constants, but it appears to me that in some cases, there is no maximum, in some cases, the maximum is 0 (when n = 1), etc. Something's seriously wrong with the question.

As for the second question i must of been watching the footy last night to much as i have rechecked my figures today and have now found a maxima point, but i am still not sure what "and find the maximum distance for the least positive value of t" means or is that just a long winded version of saying what the maxima point is?

Well, if you looked at a regular sine curve, you'd notice an infinite number of maximas. A curve such as this one will only have one global maximum, but just like the sine curve, it will have an infinite number of local maxima. Of all the local maxima and their corresponding t values, the global maximum actually occurs for the least t value, so I don't know why they bothered specifying that. Anyways, do a Yahoo! or Google search for "GCalc". The first link should take you to the GCalc page, and at the top is a link to use the GCalc applet. You should graph your function to understand what's going on. You'll notice what is meant by damped oscillation, and you'll see where the overall maximum distance occurs (as well as the local maximum distances). Actually, you see damped oscillation all the time. Hold a rope or cord or chain from your hand, and let it swing back and forth. You'll notice that the "size" of the swing slowly decreases and it eventually comes to a stop. The damped oscillation you're looking at is essentially the same (except it never comes to a stop, it continues oscillating in tiny little oscillations for ever, whereas your chain or rope will stop in real life).

 

[himanshu121]

If k,p1,p2,and n are constants then work will be constant,

there is no way diff it w.r.t n etc