A small particle is tied by the help of two strings to vertical walls with the parameters as shown in the figure $\left(g={\pi }^2\right)$. The time period of the small oscillations of the particle if it is displaced perpendicular to the plane of the paper slightly is x sec, the value of x is ____________

Sure, here are the step by step solutions: To resolve a line $P_1{P}_2$ into its parallel and perpendicular components: - Draw a line $L$ parallel to $P_1{P}_2$ passing through the point where you want to resolve it. - Draw a line $M$ perpendicular to $L$ passing through the same point. - The intersection of $P_1{P}_2$ with $L$ gives the parallel component. - The intersection of $P_1{P}_2$ with $M$ gives the perpendicular component. To show that the parallel component of $P_1{P}_2$ does not produce any torque: - Torque is produced by a force acting at a distance from the pivot point. - The parallel component of $P_1{P}_2$ acts along the line passing through $P_1$ and $P_2$, and therefore does not create any distance from the pivot point. - Therefore, the parallel component of $P_1{P}_2$ does not produce any torque. To calculate the torque produced by the perpendicular component of $P_1{P}_2$: - The torque produced by a force is given by the formula $T=rF\sin \theta$, where $r$ is the distance from the pivot point, $F$ is the force, and $\theta$ is the angle between the force and the line passing through the pivot point. - In this case, the distance $r$ is equal to the length $d$ of the perpendicular component of $P_1{P}_2$. - The force $F$ is equal to the weight of the object, which is $g_{eff}$. - The angle $\theta$ is equal to 90 degrees, since the force acts perpendicular to the line passing through the pivot point. - Substituting these values into the formula, we get $T=d{g}_{eff}\sin 90=d{g}_{eff}$. To calculate the value of $d$: - We are given the formula $T=2\pi \sqrt{\frac{d}{g_{eff}}}$. - Rearranging the formula, we get $d=\frac{T^2{g}_{eff}}{4{\pi }^2}$. - Substituting the given values, we get $d=\frac{(2\pi \sqrt{\frac{d}{g_{eff}}}{)}^2{g}_{eff}}{4{\pi }^2}=\frac{d{g}_{eff}}{\left(g\frac{\sqrt{3}}{2}\right)}$. - Simplifying this equation, we get $d=\frac{\sqrt{3}}{2}$. Therefore, the torque produced by the perpendicular component of $P_1{P}_2$ is $T=d{g}_{eff}=\frac{\sqrt{3}}{2}{g}_{eff}$, and the value of $d$ is $\frac{\sqrt{3}}{2}$.