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colonel
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Hi, in a simple spring-mass system consisting of two identical springs, how would you treat the springs? Would Hookes equation be F = kx + kx?
HallsofIvy said:Tides point is that if you attach two identical springs end to end, effectively you still have just one spring (of twice the length) still with spring constant k.
If you have two identical springs side by side (and both attached to the mass) then the act identically and then you can add them.
What if the springs aren't hanging but are placed horizontally?robphy said:Let us hang a mass M from spring 1: so, k1 x1=Mg.
Let us hang a mass M from spring 2: so, k2 x2=Mg.
If we hang the mass M from the springs arranged in series,
we have an effective spring with spring constant K and displacement X=x1+x2.
Since KX=Mg, we find
x1 = KX/k1 and
x2 = KX/k2.
A spring-mass system is a physical system that consists of a mass attached to one or more springs. The mass is able to move freely in one dimension due to the elasticity of the springs.
In a spring-mass system, the springs are treated as ideal springs, meaning they have no mass and obey Hooke's law. This means that the force exerted by the spring is directly proportional to the displacement of the mass from its equilibrium position.
The equilibrium position in a spring-mass system is the point where the spring is neither compressed nor stretched, and the mass is at rest. At this point, the forces acting on the mass are balanced, resulting in no net force.
The mass in a spring-mass system affects the frequency of oscillation and the amplitude of the motion. A larger mass will result in a lower frequency and smaller amplitude, while a smaller mass will result in a higher frequency and larger amplitude.
In a spring-mass system with two identical springs, the springs are treated as having the same spring constant and being connected in parallel. This means that the total force exerted by the springs on the mass is equal to the sum of the individual forces exerted by each spring.