Finding the period with mass attached to two springs

Click For Summary
SUMMARY

The discussion centers on deriving the period of simple harmonic motion for a mass connected to two springs with force constants k1 and k2. The established formula for the period is T = 2π√(m(k1 + k2)/(k1k2)). The user successfully identifies that when springs are connected in series, their equivalent spring constant is given by k = k1k2/(k1 + k2). This understanding is crucial for correctly applying Hooke's law and calculating the total displacement of the mass.

PREREQUISITES
  • Understanding of Hooke's Law
  • Familiarity with simple harmonic motion equations
  • Knowledge of spring constants and their behavior in series
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the equivalent spring constant for springs in series
  • Learn about the implications of mass-spring systems in physics
  • Explore the concept of damping in harmonic motion
  • Investigate the effects of varying spring constants on oscillation periods
USEFUL FOR

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

aal0315
Messages
41
Reaction score
0

Homework Statement


A mass m slides on a frictionless horizontal surface, connected to two springs. If the springs have force constants k1 and k2, show that the simple harmonic sliding motion has period: T = 2pi*sq root(m(k1+k2)/(k1k2)
There is a diagram and the springs are connected horizontally to each other and then attached to the mass.


Homework Equations


T=1/f
f=1/2pi*sq root(k/m)


The Attempt at a Solution


I figured out that for one spring T=2pi*sq root(m/k1), but i don't understand how to get k1k2 at the bottom of the equation that i need to show.
 
Physics news on Phys.org
When the two springs of spring constant k1 and k2 are connected in series, they behave like a single spring of equivalent spring constant k = k1*k2/(k1+k2).
 
This is my first post, so I hope this is okay to post! Kindly let me know if it is not.

I just did this in my homework this past weekend!

starting with Hooke's law, F=kx

for both k values:

F=k1x1
F=k2x2

so

x1=F/k1
x2=F/k2

distance traveled is x = x1+x2

x1 + x2 = F/k1 + F/k2

get common denominator of k1k2

x = (Fk2 + Fk1)/ k1k2

factor out F

x = F (k2 + k1) / k1k2

divide by F

x/F = (k2 + k1)/k1k2

now you have your k value that you can plug into your formula.
 
  • Like
Likes   Reactions: EEristavi, David89, ajcoelho and 1 other person

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 56 ·
2
Replies
56
Views
4K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
14K
Replies
20
Views
1K
  • · Replies 17 ·
Replies
17
Views
1K
Replies
17
Views
2K
Replies
5
Views
3K
Replies
6
Views
986