if a spring is cut in half what happens to its spring constant
Spring constant and stiffness constant
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Homework Argument
A mass m hangs from a spring with stiffness constant thou. The spring is cut in one-half and the same mass hung from it. Volition the new arrangement have a higher or a lower stiffness constant than the original spring?
Homework Equations
F= -kx
The Attempt at a Solution
I think that the spring will have the same spring stiffness. I dont experience that the spring constant is something that can be changed. If it was, then it shouldnt be called a constant.
Answers and Replies
Homework Argument
A mass one thousand hangs from a spring with stiffness constant one thousand. The spring is cut in half and the same mass hung from it. Will the new arrangement have a higher or a lower stiffness constant than the original spring?
Homework Equations
F= -kx
The Try at a Solution
I think that the bound will have the same spring stiffness. I dont feel that the leap abiding is something that can exist changed. If it was, then it shouldnt exist called a abiding.
as far as you lot're concerned yes it doesn't change. also as far equally i'thousand concerned it doesn't considering i'm a noob but i exercise know that information technology does actually change because i know that cut intermission springs changes their frequency of oscillation which is directly proportional to the root of [tex]\frac{M}{m}[/tex] which means the M does change. i just don't know how
On a side annotation, I think my arrow keys are disable in the forum and when I hit the apostrophe cardinal I become a quick search link. Is anyone else having the same problem?
No, it is the bound constant. It does not change assuming it is the same spring. It will just one-half the amount of compression that the jump tin can take before you go past the limit.On a side notation, I call back my pointer keys are disable in the forum and when I hit the apostrophe key I get a quick search link. Is anyone else having the same problem?
yup yous're right, the cut of the springs affecting springs rate is considering it messes upward the local annealing in the bound
yup you're right, the cutting of the springs affecting springs rate is because it messes upward the local annealing in the spring
I don't really empathize what y'all only said. I as well don't understand what is meant by "rate".
I don't really sympathise what you just said. I also don't sympathise what is meant past "rate".
free frequency of oscillation is what i meant past leap rate,
Damage to springs, such as using oxy-acetylene to cut the end off a auto suspension spring to lower a vehicle's ride height, can destroy the tempering in localised areas of the spring.
http://en.wikipedia.org/wiki/Coil_spring
tempering i meant the tempering of the spring is affected
If it isn't a car spring I highly incertitude that you would utilise a acetylene torch to cutting it. Say a mechanical pencil spring
you were right, i was incorrect, i was dislocated virtually why the K of the spring is afflicted.
i didn't think you lot came off as condescending
Think nearly the definition of spring abiding. If a bound is cutting in half, is information technology easier or harder to stretch it a given corporeality?I recollect that the spring volition have the same spring stiffness.
Some other mode to look at it: Hang a weight from the original spring and it stretches the spring a altitude X. Now imagine that the spring is really two springs attached together (each i half the size). How much does that weight stretch each half-bound?
It's a constant for a given spring--equally long as you don't stretch it too far. In Hooke's constabulary (F = -kx), F and x alter as you stretch the spring, simply k remains constant. If yous modify the spring by cutting it, you create a new bound with a new bound constant.I dont experience that the spring abiding is something that can exist inverse. If information technology was, so it shouldnt be called a abiding.
Think about the definition of bound constant. If a bound is cut in one-half, is information technology easier or harder to stretch information technology a given amount?Another way to await at it: Hang a weight from the original bound and it stretches the bound a distance X. At present imagine that the spring is really 2 springs fastened together (each one half the size). How much does that weight stretch each half-jump?
Information technology'southward a constant for a given spring--equally long as yous don't stretch information technology likewise far. In Hooke'south law (F = -kx), F and x change equally you stretch the spring, merely one thousand remains constant. If you lot change the bound by cutting it, you create a new spring with a new spring constant.
what is the definition of the spring constant? is it non F/x
That'southward correct.what is the definition of the jump constant? is it not F/10
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