what is the physical significance of k/m?
this question is related to uniform circular motion and centripetal force..
I don`t understand the question,and what are the points should I include while answering this question.Help me please!
- gintableLv 71 decade agoFavorite Answer
In the above poster's description, k/m only has significance in a situation which has nothing to do with uniform circular motion...ok, maybe a little, because they are both situations of periodic motion.
k/m means the stiffness constant of a spring (k) divided by the mass (m) of the attached object. The cycles per second frequency in such a situation is found by f = sqrt(k/m)/(2*Pi).
As for uniform circular motion...have you tried just posting an example from your homework or textbook verbatim...and then post what parts you don't understand? That works for most people.
Also, in uniform circular motion...remove the terms "centripetal force", "centrifugal force", and "Coriolis force" all from your vocabulary. They are the most unnecessary and confusing concepts in introductory physics. Use the word "centripetal" ONLY as an adjective with the word "acceleration".
Instead think about what those force terms really mean:
Instead of saying "centripetal force", call it the "net force needed to cause centripetal acceleration". It could be tension, friction, normal force, gravity, human force, magnetism, or a combination of forces which result in the total force needed to cause centripetal acceleration.
Instead of saying "centrifugal force", say "centrifugal effect". What this really is, is an illusion within a rotating environment, that gives occupants the illusion that an "equivalence of gravity" is causing them to flee the center of the circular motion. It is nothing more than a term which appears in a construction of Newton's laws in a rotating reference frame. It isn't a real force.
Same is true with "Coriolis force"...IT'S the "Coriolis EFFECT". Again, it is an illusion to occupants of a rotating environment and it is nothing more than a term which appears in a construction of Newton's laws in a rotating reference frame. It isn't a real force.
- kirchweyLv 71 decade ago
In a mass-spring system, k/m determines the natural frequency: w = sqrt(k/m) rad/s.
Perhaps you could supply a little more context? What is k?