I'm a Fruit Ninja (literally)

The diagram on the left shows the initial state of the system. Since it is not mentioned in the question, I am assuming that both springs are initially undeformed. Assuming that the natural length of the spring is $L_o$ . The diagram on the right shows the state of the system when the rod rotates by an angle $\theta$ at a general time $t$ .

The coordinates of each end of the spring at this general instant are as follows. The ends of each spring fixed to the walls have coordinates $(L_o, L/2)$ and $(-L_o,-L/2)$ . The ends of each spring fixed to the rod have coordinates:

$\left(\frac{L\sin{\theta}}{2},\frac{L\cos{\theta}}{2}\right)$ $\left(-\frac{L\sin{\theta}}{2},-\frac{L\cos{\theta}}{2}\right)$

With this information, we can compute the sum of potential energies of both springs. This comes out to be:

$V = 2\left(\frac{K}{2} \left(L_S - L_o\right)^2 \right)$

$L_S=\left( \left(L_o -\frac{L\sin{\theta}}{2}\right)^2 + \left(\frac{L}{2}-\frac{L\cos{\theta}}{2}\right)^2 \right)^{0.5}$

Since we are allowed to assume that $\theta$ is small, then: $\sin{\theta} \approx \theta$ and $\cos{\theta} \approx 1$ . This simplifies the potential energy expression to:

$V = \frac{KL^2\theta^2}{4}$

At the same instant, the kinetic energy of the rod as it rotates about the origin is:

$T = \frac{1}{2} \left(\frac{mL^2}{12}\right) \dot{\theta}^2$

Since there are no dissipative forces acting on the system, the total energy of the system is constant:

$E = T + V$ $E = \frac{mL^2\dot{\theta}^2}{24} + \frac{KL^2\theta^2}{4}$ $\frac{dE}{dt} = 0 = \left(\frac{m\ddot{\theta}}{12} + \frac{K\theta}{2} \right) L^2 \dot{\theta}$

The above is only possible when:

$\frac{m\ddot{\theta}}{12} + \frac{K\theta}{2} = 0$

Which is the required equation of motion (EOM). Since $m = 1.5$ , this simplifies the EOM to:

$\ddot{\theta} + 4K \theta = 0$

The above equation implies that small oscillations are simple harmonic in nature. This means:

$\implies 2 \pi f = 2 \sqrt{K}$ $\boxed{f = \frac{\sqrt{K}}{\pi}}$

when the sword is displaced

Formulae used :

• $\tan\theta=\dfrac{perpendicular}{base}$

• torque $(\tau)=K\theta$

• torque $(\tau) =I \alpha$ where $\alpha$ is the angular acceleration and I is the moment of inertia

• $f=\dfrac{\omega}{2π}$

• moment of inertia of a rod = $\dfrac{ML^{2}}{12}$

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Here $\tan\theta = \dfrac{x}{\tfrac{L}{2}}$ Since $\theta$ is small, $\tan\theta≈\theta$

$=>x=\dfrac{L{\theta}}{2}$

The restoring torque acting on the sword when it is displaced from its initial position:

$\tau =-2kx*\dfrac{L}{2}=-KxL$

Angular acceleration $\alpha = \dfrac{\tau}{I} =\dfrac{-K \tfrac{L\theta}{2} L}{I}$

$=> \alpha = \dfrac{-6K\theta}{mass} = - \omega^{2} \theta$

Comparing the two equations we get ,

$\omega =\sqrt{\dfrac{6k}{\tfrac{6}{4}}} = 2 \sqrt{K}$

$=> frequency=\dfrac{2\sqrt{K}}{2π}$

$=>\boxed{ frequency=\dfrac{\sqrt{K}}{π}}$