Mass attached to lower of 2 identical springs

In summary, for a mass hanging from a uniform spring with spring constant k, the period of oscillations is T=2π√(m/k). If the mass is hung from two identical springs connected side by side, the period becomes T=2π√(m/2k). If the mass is hung from the bottom of two identical springs connected end to end, the period is T=2π√(m/(k/2)). The stiffness of the combined springs in this case is 2k.
  • #1
bfusco
128
1

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


[itex]T=2\pi /\omega [/itex]

The Attempt at a Solution


(a)[itex]m\ddot{x} +kx=0[/itex]

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

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

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

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

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

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

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

(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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  • #2
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.
 
  • #3
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".
 
  • #4
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?
 
  • #5
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=?
 
  • #6
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?
 
  • #7
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.
 
  • #8
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
 

What is a mass attached to lower of 2 identical springs?

A mass attached to lower of 2 identical springs refers to a physical system where a mass is connected to two identical springs, one on top and one on the bottom. This system is often used in physics experiments to study the behavior of springs and their relation to mass.

How do the two identical springs affect the mass?

The two identical springs work together to support the mass and determine its movement. When the mass is pulled down, the springs will stretch and apply a restorative force to bring the mass back to its equilibrium position. The stiffness of the springs will also determine the frequency of the oscillations of the mass.

What is the equation for the motion of a mass attached to lower of 2 identical springs?

The equation for the motion of a mass attached to lower of 2 identical springs is given by: F = -kx - mg, where F is the net force on the mass, k is the spring constant, x is the displacement of the mass from its equilibrium position, and mg is the force due to gravity.

How does the mass affect the motion of the system?

The mass affects the motion of the system by determining the amplitude of the oscillations. A larger mass will have a larger amplitude, meaning it will oscillate with a greater range of motion. The mass also affects the frequency of the oscillations, with a larger mass resulting in a lower frequency.

What factors can affect the behavior of a mass attached to lower of 2 identical springs?

There are several factors that can affect the behavior of a mass attached to lower of 2 identical springs. These include the mass of the object, the stiffness of the springs, the amplitude of the oscillations, and any external forces acting on the system. Other factors such as temperature and damping can also affect the behavior of the system.

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