- #1
JoeyBob
- 256
- 29
- Homework Statement
- See attached
- Relevant Equations
- Shown in body
So the textbook uses 3 equations for these,
d 2/dt2 (y1 + y3 − √ 2y2) = − k / m (2 + 1/ √ 2 ) (y1 + y3 − √ 2y2)
d 2 /dt2 (y1 + y3 + √ 2y2) = − k / m (2 − 1/ √ 2 ) (y1 + y3 + √ 2y2)
d 2/ dt2 (y1 − y3) = − 2k / m (y1 − y3)
Now the question is asking for the largest natural frequency. Now I must be applying the above equations the wrong way, but I chose number 1 of the above where = − k m (2 + 1 / √ 2 ) (y1 + y3 − √ 2y2).
What I did to find w was make it equal the part in front of the y1, y2 ect like this
w=sqrt(k/(2m)+k/(2m sqrt(2))). Now this gave me the wrong answer of 2.07 when the answer is suppose to be 2.32, quite a bit higher. The next question also asks the same thing but for the smallest frequency and if I use equation 3 of the above in the same way to get w, I will also get the wrong answer.
d 2/dt2 (y1 + y3 − √ 2y2) = − k / m (2 + 1/ √ 2 ) (y1 + y3 − √ 2y2)
d 2 /dt2 (y1 + y3 + √ 2y2) = − k / m (2 − 1/ √ 2 ) (y1 + y3 + √ 2y2)
d 2/ dt2 (y1 − y3) = − 2k / m (y1 − y3)
Now the question is asking for the largest natural frequency. Now I must be applying the above equations the wrong way, but I chose number 1 of the above where = − k m (2 + 1 / √ 2 ) (y1 + y3 − √ 2y2).
What I did to find w was make it equal the part in front of the y1, y2 ect like this
w=sqrt(k/(2m)+k/(2m sqrt(2))). Now this gave me the wrong answer of 2.07 when the answer is suppose to be 2.32, quite a bit higher. The next question also asks the same thing but for the smallest frequency and if I use equation 3 of the above in the same way to get w, I will also get the wrong answer.