As shown in the figures,a uniform rod OO^{pri | vedclass

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(C) For Fig. $1$,the restoring torque $	au$ for a small angular displacement $	heta$ is given by the sum of torques from both springs. The spring at

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$\sqrt{\frac{5}{2}}$

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(C) For Fig. $1$,the restoring torque $	au$ for a small angular displacement $	heta$ is given by the sum of torques from both springs. The spring at the midpoint is at a distance $l/2$ from the hinge $O$,and the spring at the top end is at a distance $l$ from the hinge $O$. The restoring torque is $	au = -(K \cdot (l/2	heta) \cdot l/2 + K \cdot (l	heta) \cdot l) = -K	heta(l^2/4 + l^2) = -\frac{5}{4}Kl^2	heta$.Using the equation of motion for rotation,$I\alpha = 	au$,where $I = \frac{Ml^2}{3}$ is the moment of inertia about the hinge $O$ and $\alpha = \ddot{	heta}$:$\frac{Ml^2}{3} \ddot{	heta} = -\frac{5}{4}Kl^2	heta \implies \ddot{	heta} + \frac{15K}{4M}	heta = 0$.The angular frequency is $\omega_1 = \sqrt{\frac{15K}{4M}}$.For Fig. $2$,both springs are connected at the midpoint $(l/2)$. The restoring torque is $	au = -(K \cdot (l/2	heta) \cdot l/2 + K \cdot (l/2	heta) \cdot l/2) = -2K(l/2)^2	heta = -\frac{1}{2}Kl^2	heta$.Using the equation of motion,$\frac{Ml^2}{3} \ddot{	heta} = -\frac{1}{2}Kl^2	heta \implies \ddot{	heta} + \frac{3K}{2M}	heta = 0$.The angular frequency is $\omega_2 = \sqrt{\frac{3K}{2M}}$.The ratio of frequencies is $\frac{f_1}{f_2} = \frac{\omega_1}{\omega_2} = \sqrt{\frac{15K}{4M} \cdot \frac{2M}{3K}} = \sqrt{\frac{15}{6}} = \sqrt{\frac{5}{2}}$.

Column-$I$	Column-$II$
$(A)$ Velocity-displacement graph $(\omega \neq 1)$	$(i)$ Straight line
$(B)$ Acceleration-displacement graph	$(ii)$ Sinusoidal
$(C)$ Acceleration-time graph	$(iii)$ Circle
$(D)$ Acceleration-velocity graph $(\omega \neq 1)$	$(iv)$ Ellipse

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