Frequency of oscillation of a rod with two springs

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Let's take that in equilibrium, both springs S_l and S_r are elongated by $x_e$.
Assuming that the oscillations are so small that the springs could be taken along x- axis approximately.
At time t, S_l is elongated by x. The rod makes an angle $\theta$ with vertical axis such that
$\tan \theta = \frac x {\frac l 2 }$
This gives that the S_r will be elongated by x' = $2 \tan \theta = 2x$

Eqn. of motion of com gives,
$k ( x_e + x ) (-\hat x ) +k (x_e -2x) \hat x = m \ddot x \hat x$
$-3kx = m\ddot x$
$\omega = \sqrt (\frac {3k} m )$

Is this correct so far?

 

[Pushoam]

Another attempt:
Motivated from the OP's solution,
Torque eqn. about the pivot-point,
$k ( x_e + x ) \frac l 2 (-\hat z ) +k (x_e -2x) l \hat z= I \ddot \theta \hat z$
But in equilibrium, the above eqn. gives me non-zero net torque, so it means that the elongation is not same for both the springs.
Let's say that in equilibrium, LHS elongation is $x_{le}$ and RHS elongtion is $x_{re}$, and $x_{le } ≠ x_{re}$ . But these elongations are such that the net torque on the system about the pivot is zero in the equilibrium
But, this tells that net force acting on the com is non - zero in equilb. { A CONTRADICTION ?}

$\tan \theta = \frac {2x } l \\ \text{ for small , } \theta , \\ \tan \theta≈ \theta \ddot \theta = \frac {2 \ddot x} l$
So,
$k x \frac l 2 (-\hat z ) +k ( -2x) l \hat z= I \ddot \theta \hat z = \frac {ml^2}3 (\frac {2 \ddot x} l)$
$\frac { -5} 2 kx = \frac 2 3 m \ddot x$
$\omega = \sqrt {(\frac {15 k} {4 m}) }$

But how to decide the correctness of both the solutions?

 

[haruspex]

in equilibrium, both springs S_l and S_r are elongated by x_e.

Is that possible? What would the two torques about the pivot be?

 

[Pushoam]

Is that possible?

No, it is not.
In the OP, I thought that it were only springs which were exerting forces along x-axis. Consequently, forces by them should be equal in magnitude in eqbm and so their elongation.
But, torque analysis in post #2 says that it is not so.
SO, I think the HINGE also exert force along x-axis to make the center of mass in eqbm. I missed it.

LEARNING: IN CASE OF RIGID BODY MOTION, I SHOULD DO BOTH TORQUE AND FORCE ANALYSIS BEFORE GOING AHEAD.
So, the solution in post #2 is correct.
Is this correct so far?

 

[haruspex]

the solution in post #2 is correct.

Yes.

 

[Pushoam]

Thanks.