Mass attached to lower of 2 identical springs

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Homework Help Overview

The discussion revolves around the oscillation periods of a mass attached to springs, specifically focusing on a single spring and configurations involving two identical springs. Participants explore the implications of spring constants and the setup of the system, questioning how to approach the calculations for different arrangements of the springs.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the period of oscillations for a mass on a single spring and for two springs in parallel and in series. There are attempts to derive the angular frequency and period using differential equations. Questions arise regarding the correct setup for the spring constants in different configurations and the impact of the mass's weight.

Discussion Status

Some participants have provided guidance on analyzing the system's equilibrium and the forces acting on the springs. There is ongoing exploration of how to combine the spring constants and the total stiffness of the system. Multiple interpretations of the spring configurations are being considered, and participants are actively questioning their assumptions and calculations.

Contextual Notes

Participants express uncertainty about neglecting the weight of the mass and how it affects the system. There are references to specific equations and concepts from their textbooks, indicating a reliance on established principles while grappling with the nuances of the problem.

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Homework Statement


a mass m hangs from a uniform spring of spring constant k.
(a)what is the period of oscillations in the system?
(b)what would it be if the mass m were hung so that:
1)it was attached to two identical springs hanging side by side?
2)it was attached to the lower of two identical springs connected end to end?

(P.S. I am not sure if we are to neglect the force of weight of the mass)

Homework Equations


T=2\pi /\omega

The Attempt at a Solution


(a)m\ddot{x} +kx=0

dividing by "m" you get: \ddot{x}+\omega^2 x=0

to find the roots: r^2 +\omega^2 = 0
to which the roots are: r=\pm i\omega. According to this equation the angular frequency is \omega which equals \sqrt{k/m}.

Therefore, T=2\pi \sqrt{m/k}

(b)(1)skipping a few steps because its similar to last equation you get \ddot{x} +\omega^2 x=0

again finding the roots you get: \pm \omega \sqrt{2} i,

According to this equation the angular frequency is \sqrt{2}\omega which equals \sqrt{2k/m}.

the period is: 2\pi \sqrt{m/2k}

(b)(2) I am not really sure how to set this one up but my thinking was something like this.
I can write the spring force as if it were 1 spring of length 2L.

but the thing is that the oscillations don't depend on the length of the spring, and I am not sure if this is a correct way of going about it. Please help.
 
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For part 2, you need to figure out the stiffness of the spring combo. You could examine the static case (equilibrium) and draw the FBD.
 
voko said:
For part 2, you need to figure out the stiffness of the spring combo. You could examine the static case (equilibrium) and draw the FBD.

Right i did that, however i don't know how to combine the spring constants "k".
 
Consider that a mass is attached to two springs (one on top of the other) of equal stiffness and the system is in equilibrium.

What is the force acting on the top spring? On the bottom spring? How much extension does each spring have? What is the total extension? What is the total stiffness?
 
voko said:
Consider that a mass is attached to two springs (one on top of the other) of equal stiffness and the system is in equilibrium.

What is the force acting on the top spring? On the bottom spring? How much extension does each spring have? What is the total extension? What is the total stiffness?

in this "s" means the unstretched length of the spring, and "1" refers to top spring, "2" refers to bottom spring.

F1=mg-k(s+x)

F2=mg-k(s+x)
where s is the natural length of spring

spring 1 extends (s+x)
spring 2 extends (s+x) as well, only from the bottom of the top spring.
total extension=2(s+x)
total stiffness=?
 
Hooke's law is F = kx, not F = k(s + x). Drop the s'es in your equations.

Other than that, you are almost there. Treat the two springs as one spring. You know the total force it is loaded with. You have computed its total extension. What is its stiffness then?
 
voko said:
Hooke's law is F = kx, not F = k(s + x). Drop the s'es in your equations.

After a mass m is attached to a spring, it stretches the spring by an amount s and attains a position of equilibrium at which its weight mg is balanced by the restoring force ks. which results in the final equation F=-kx. This way of looking at spring equations helps me understand basic equations better because when i am summing the forces, mg is an obvious one, and i need to know how to handle it.

voko said:
Other than that, you are almost there. Treat the two springs as one spring. You know the total force it is loaded with. You have computed its total extension. What is its stiffness then?

I want to say the stiffness would just be k because if you take a spring of stiffness k, and divide it into many small springs, i believe all those little springs would have stiffness k as well. But according to my textbook that is incorrect and i don't know why. I believe my textbook says it should be k/2.
 
I figured it out. i didnt know how to take into account that only 1 spring was touching the mass. F=-k(x1+x2), F1=-k1x1, F2=-k2x2, set them equal and solve for x2, plug into F=-k(x1+x2), and you get 1/k=1/k1+1/k2
 

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