**1. The problem statement, all variables and given/known data**

We have two springs each ##0.5 m## long and both have ##k=5N/cm##. One end of the spring is attached to the wall while the other two ends are connected in the middle. On the junction point of the two springs a mass is hanged. How does the frequency of oscillation around equilibrium state depend on the initial offset due to the mass?

Hint: There is no need to explicitly write the initial offset as function of mass. Just use the offset as independent parameter using which mass can be calculated.

**2. Relevant equations**

**3. The attempt at a solution**

##mg+2Fsin\varphi=ma##

##mg+2k\frac{z}{\sqrt{l^2+z^2}}(\sqrt{l^2+z^2}-l)=m\ddot{z}##

Let’s say that ##z=D+\varepsilon ## where ##\varepsilon ## is very small.

##mg+2k\frac{D+\varepsilon }{\sqrt{l^2+(D+\varepsilon )^2}}(\sqrt{l^2+(D+\varepsilon )^2}-l)=m\ddot{\varepsilon }##

##mg+2k(D+\varepsilon)-\frac{2kl(D+\varepsilon )}{l\sqrt{1+(\frac{D+\varepsilon }{l})^2}}=m\ddot{\varepsilon }##

##mg+2k(D+\varepsilon )-2k(D+\varepsilon)[1-\frac 1 2 (\frac{D+\varepsilon }{l})^2]=m\ddot{\varepsilon }##

##mg+\frac{2k}{l^2}(D+\varepsilon )^3=m\ddot{\varepsilon }##

##mg+\frac{2kD^3}{l^2}[1-3\frac \varepsilon D]=m\ddot{\varepsilon }##

##\ddot{\varepsilon }+\frac{6kD^2}{ml^2}\varepsilon =g+\frac{2kD^3}{l^2}##

So if I am not mistaken, than ##\omega \propto D##. Right?

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