Finding the period with mass attached to two springs

AI Thread Summary
The discussion focuses on deriving the period of a mass attached to two springs with spring constants k1 and k2. The key equation for the period of simple harmonic motion is T = 2π√(m(k1 + k2)/(k1k2)). The user initially calculated the period for a single spring but needed clarification on how to combine the two spring constants. It was explained that when springs are connected in series, their equivalent spring constant is given by k = k1k2/(k1 + k2). The user successfully demonstrated the derivation using Hooke's law and the relationship between the forces and displacements of the springs.
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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.
 
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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.
 
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