Mass attached to lower of 2 identical springs

  • Thread starter Thread starter bfusco
  • Start date Start date
  • Tags Tags
    Mass Springs
Click For Summary
SUMMARY

The discussion focuses on the oscillation periods of a mass attached to springs with different configurations. For a single spring with spring constant k, the period of oscillation is T = 2π√(m/k). When the mass is attached to two identical springs in parallel, the effective spring constant doubles, resulting in a period of T = 2π√(m/2k). In the case of two springs in series, the effective spring constant is halved, leading to a period of T = 2π√(2m/k). The participants clarify the importance of correctly combining spring constants and understanding the equilibrium conditions for accurate calculations.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Basic knowledge of oscillatory motion and angular frequency
  • Familiarity with differential equations related to harmonic motion
  • Ability to analyze forces in static equilibrium
NEXT STEPS
  • Study the derivation of oscillation periods for systems with multiple springs
  • Learn about effective spring constants in parallel and series configurations
  • Explore the principles of static equilibrium in mechanical systems
  • Investigate the applications of harmonic motion in real-world scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory systems, as well as educators seeking to clarify concepts related to spring dynamics and harmonic motion.

bfusco
Messages
126
Reaction score
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


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.
 
Physics news on Phys.org
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
 

Similar threads

Replies
17
Views
3K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
4
Views
3K
Replies
31
Views
2K
Replies
7
Views
1K
Replies
6
Views
988
Replies
3
Views
1K