Expert: Ahmed Salami - 11/11/2004

Question
Ok, so i got the following Quadratic Equation:

m1m2x^2 + m1K2x + m1K3X + m2xK1 + k1k2 + k1k3 + k2M2x + k2k3
___________________________________________________________
         m1m2

x^2 + (m1k2 + m1k3 + m2k1 + m2k2)x + (k1k2 + k1k3 + k2k3) = 0
      -------------------------      ------------------
         m1m2          m1m2

Now, using


x = -b +- root (b2-4ac) / 2a to get x... i am not sure how to obtain the two values when there are more than one constant (k1 m1 m2 etc)....



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Followup To
Question -
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Followup To
Question -
I am having trouble understanding how i can show taht the following 2x2 matrix has two distinct eigenvalues that are both negative

Ok, thanks.... i got the following

m1m2x^2 + m1K2x + m1K3X + m2xK1 + k1k2 + k1k3 + k2M2x + k2k3
___________________________________________________________
         m1m2

for the characteristic polynomial. I am unsure how to obtain the form of the quadratic equation from the above and obtain the two eigenvalues from that....


Thanks.












A =  -(k1+k2)/m1          k2/m1
    k2/m2          -(k2+k3)/m2


I can obtain the characteristic polynomial but cannot simplify it to show that it has two distinct negative eigenvalues


I am told that k1, k2, k3 are >0 which represent the spring constants for the question. Also, m1 and m2 are masses .: positive
Answer -
Hi K,
The characteristic polynomial would be a quadratic in #(lambda). With the knowledge of the fact that k1,k2,k3,m1,m2 are >0 , the equation can be considered to be of the form
#^2 + p# + q = 0
where p and q are positive
Hence, it is clear that only a negative value of # would satisfy the equation because #^2 is always positive and only a negative # would be able to neutralize the effects of all the positives.
But as a little proof, solving this equation gives
# = [-p +or- sqrt(p^2 - 4q)]/2
[-p - sqrt(p^2 - 4q)]/2 would always be negative.
[-p + sqrt(p^2 - 4q)]/2 would only be positive if
sqrt(p^2 - 4q) > p   i.e
(p^2 - 4q) > p^2
or simply 4q < 0 which appears to be a contradiction since q > 0.
I'll be here to explain anything unclear.
Regards.
Answer -
Hi K,
From where you stopped, all you need do is just make a simple rearrangement.
First divide through by m1m2 and collect all terms in x. The coefficient of x^2 would obviously be 1(i.e m1m2/m1m2). Add all the coefficients of x to get the p i mentioned in the first reply. The remaining numbers would be the constant q. Considering that the m's and k's are positive, you'll surely see that p and q would be positive.
I just want you to do it on your own. but if you still have problems, let me know.
Good luck.
Regards.

Answer
Hi K,
Sorry about the delay.
Well, you really don't have to use that formula to find the roots to confirm that the two roots would be negative. Using the formula only gives you unnecessary and tedious work since the question as i can recall only asks you to show that the roots are negative.
Now, just follow the line of argument in my first reply to you which i would state again at the end of this and you'll be able to prove that the roots are in fact negative.

The characteristic polynomial would be a quadratic in #(lambda). With the knowledge of the fact that k1,k2,k3,m1,m2 are >0 , the equation can be considered to be of the form
#^2 + p# + q = 0
where p and q are positive
Hence, it is clear that only a negative value of # would satisfy the equation because #^2 is always positive and only a negative # would be able to neutralize the effects of all the positives.
But as a little proof, solving this equation gives
# = [-p +or- sqrt(p^2 - 4q)]/2
[-p - sqrt(p^2 - 4q)]/2 would always be negative.
[-p + sqrt(p^2 - 4q)]/2 would only be positive if
sqrt(p^2 - 4q) > p i.e
(p^2 - 4q) > p^2
or simply 4q < 0 which appears to be a contradiction since q > 0.
I'll be here to explain anything still unclear.

Regards.