Pull, and let go

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Let the Mass $\displaystyle M$ be displaced by a distance $\displaystyle x$ . Let the corresponding extensions in the springs be $\displaystyle y$ and $\displaystyle z$ .

It is important to note that the extensions in the springs will NOT be the same. Since the tension in the string that joins the two springs is constant, we have,

$T = k_1y = k_2z$

Also, using constraint relations, we get,

$y+z = 2x$

Solve the above two equations to get $\displaystyle y$ and $\displaystyle z$ in terms of $\displaystyle x$

Now, writing the Force equation for the mass $\displaystyle M$ ,

$2T-Mg = Ma$

$2(k_1y)-Mg = Ma$

$2k_1 (\frac{2k_2}{k_1+k_2}x)-Mg = Ma$

$a = \frac{4k_1k_2}{M(k_1+k_2)} x - g$

Therefore,

$w^2 = \frac{4k_1k_2}{M(k_1+k_2)}$

$w = \sqrt{\frac{4k_1k_2}{M(k_1+k_2)}}$

Hence, the Time period will be,

$T = \frac{2\pi}{w} = 2\pi\sqrt{\frac{M(k_1+k_2)}{4k_1k_2}}$

Substituting the values provided,

$\boxed{T = 0.76952\text{sec}}$

Another way to approach this is through the Lagrangian.

The kinetic energy is clearly $\frac12 M \dot{x}^2$ .

We have

$\begin{aligned} y+z &= 2x \\ \dot{y} + \dot{z} &= 2\dot{x} \end{aligned}$

Since the tension is given by $T = k_1y=k_2z$ , we have $\displaystyle 2x = \frac{T}{k_1} + \frac{T}{k_2}$ or $\displaystyle T = 2x\frac{k_1k_2}{k_1+k_2}$ .

The potential energy is thus given by $\displaystyle V = 2x^2\frac{k_1k_2^2 + k_2^2k_1}{\left(k_1+k_2\right)^2} - Mgx= 2x^2\frac{k_1k_2}{k_1+k_2} - Mgx$ .

The Lagrangian is thus given by

$\mathcal{L} = \frac12 M \dot{x}^2 - 2x^2\frac{k_1k_2}{k_1+k_2} - Mgx$

which yields the equation of motion

$M\ddot{x} = 4x\frac{k_1k_2}{k_1+k_2}$

Showing that $\boxed{\displaystyle\omega^2 = \frac{4}{M}\frac{k_1k_2}{k_1+k_2}}$ as expected.