Mechanical Vibration-differential eq Assume a 8 lb weight stretches a spring 6 i
ID: 1780889 • Letter: M
Question
Mechanical Vibration-differential eq
Assume a 8 lb weight stretches a spring 6 inches. Recall that the gravity constant for standard units is 32ft/s232ft/s2 (also recall that a pound is a unit of force (weight) not mass). Suppose we also consider the effect of dampening, where when the mass is moving 2ft/sec2ft/sec the medium in which the mass moves imparts of force of 8 lbs. Is this system overdamped, critically damped, or underdamped? Justify with the appropriate computations. (You do not need to find a solution; Please just classify the system).
SOLUTION:
The mass is (1/4) slugs, and the spring constant is 8=k/2 so k=16k=16. The dampenening constant is 4. Then the equation is u/4+4u+16u=0u/4+4u+16u=0. The only root to the characteristic equation is -8, so the system is critically dampened.
MY QUESTION: I am not very convinced with 8=k/2, can you please explain, since hook's law states that F=kx, but according to the solution in the book, it is f=k/x or what that 2 is, and where is that from.
Explanation / Answer
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as per hooke's law, F=k*x
here F=8 lb and x =6 inches=(6/12)=1/2 foot
hence 8=k*(1/2)=k/2
==>k=16 lb/ft
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the unit should be in sync and if you are using lb as unit for weight, correct unit for for length is foot and 12 inches=1 ft , hence 6 inches=1/2 ft