A classical mechanics problem by Godwin Tom George

First of all, Clearly, since the use of springs is linearly involved, we can derive that the motion shown by the bar, is a an SHM. Now, in a Simple Harmonic Motion, the total energy of the body remains constant throughout the motion. So, what I'm going to do, is to take an instant, and differentiate the total energy of the system with respect to displacement, and equate it to $0$ .

Let at an instant, the extension in both the springs is $'x'$ each, as the situation is symmetrical. Let, the mass of the body be $'M'$ and it's length be $'l'$ . Let at this instant the angular velocity of the body be $\omega$ . So, total energy at this instant is : $Total\quad Energy\quad (E)\quad =\quad \frac { 1 }{ 2 } K{ x }^{ 2 }+\frac { 1 }{ 2 } K{ x }^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 }$ , where $I$ is the moment of inertia of the bar.

Differentiating this expression with respect to $'x'$ : $\quad \quad \frac { dE }{ dx } \quad =\quad 0\\ \Rightarrow \frac { d }{ dx } (\frac { 1 }{ 2 } K{ x }^{ 2 }+\frac { 1 }{ 2 } K{ x }^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 })\quad =\quad 0\\ \Rightarrow \frac { d }{ dx } (K{ x }^{ 2 })\quad +\quad \frac { d }{ dx } (\frac { 1 }{ 2 } I{ \omega }^{ 2 })\quad =\quad 0\\ \Rightarrow \frac { d }{ dx } (-K{ x }^{ 2 })\quad =\quad \frac { d }{ dx } (\frac { 1 }{ 2 } { I\omega }^{ 2 })\\ \Rightarrow \quad -2Kx\quad =\quad \frac { 1 }{ 2 } \frac { d }{ dx } (\frac { M{ l }^{ 2 } }{ 12 } { \omega }^{ 2 })\\ \Rightarrow \quad -4Kx\quad =\quad \frac { M{ l }^{ 2 } }{ 12 } \frac { d }{ dx } (\frac { { v }^{ 2 } }{ { (\frac { l }{ 2 } ) }^{ 2 } } )\\ \Rightarrow \quad -4Kx\quad =\quad \frac { M{ l }^{ 2 } }{ 12 } \frac { 4 }{ { l }^{ 2 } } \frac { d }{ dx } ({ v }^{ 2 })\\ \Rightarrow \quad -4Kx\quad =\quad \frac { M }{ 3 } 2v\frac { d }{ dx } v\\$

But, $v\frac { dv }{ dx } \quad =\quad a$

So, $\Rightarrow \quad -6Kx\quad =\quad Ma\\ \Rightarrow \quad F\quad =\quad -6Kx$

Clearly , this is an SHM with spring factor $K'$ = $6K$

Now, time period for an SHM (T) = $2\pi \sqrt { \frac { Inertial\quad Factor }{ Spring\quad Factor } }$ Here, Inertial factor is the mass of the body, and the spring factor is $K'$ = $6K$

So, $\quad \quad \quad T\quad =\quad 2\pi \sqrt { \frac { M }{ K' } } \\ \Rightarrow \quad T\quad =\quad 2\pi \sqrt { \frac { 6 }{ 6K } } \\ \Rightarrow \quad T\quad =\quad 2\pi \sqrt { \frac { 6 }{ 6 } } \\ \Rightarrow \quad T\quad =\quad 2\pi \quad seconds\quad =\quad 6.28s$