Engineering Mechanics

As section is symmetric about given y-y axis, hence centroid will lie on this axis.
To calculate location of C.G. from reference axis (bottom horizontal line); \bar{y}
\bar{y}=\frac{A_{1} \bar{y}_{1}+A_{2} \bar{y}_{2}}{A_{1}+A_{2}}
\bar{y}=\frac{(40 \times 10) \times 55+(20 \times 50) \times(25)}{40 \times 10+20 \times 50}
\bar{y}=33.5714 \mathrm{~cm} (from bottom)
MOI of section about x x (centroidal horizontal axis)
I=I_{I, x x}+I_{I I, x x}
I_{1, x x}=\frac{b_{1} d_{1}^{3}}{12}+\left(b_{1} d_{1}\right) \times d^{2}

\begin{aligned} & \mathrm{I}_{1, x x}=\frac{40 \times 10^{3}}{12}+(40 \times 10)(55-33.57)^{2} \\ & \mathrm{I}_{1, x x}=187031.29 \mathrm{~cm}^{4} \end{aligned}

\mathrm{d}^{\prime}=\overline{\mathrm{y}}-25
\mathrm{d}^{\prime}=8.5714 \mathrm{~cm}
\mathrm{I}_{\mathrm{II}, x x}=\frac{\mathrm{b}_{2} \mathrm{~d}_{2}^{3}}{12}+\left(\mathrm{b}_{2} \times \mathrm{d}_{2}\right) \times \mathrm{d}^{\prime 2}
I_{I I, x x}=\frac{20 \times 50^{3}}{12}+(20 \times 50) \times(8.57)^{2}
\mathrm{I}_{\mathrm{II}, \mathrm{xx}}=281778.23 \mathrm{~cm}^{4}

Hence, MOI of whole section
\mathrm{I}_{x \mathrm{x}}=\mathrm{I}_{\mathrm{I}, x \mathrm{x}}+\mathrm{I}_{\mathrm{Il}, \mathrm{xx}}
\mathrm{I}_{x x}=468809.52 \mathrm{~cm}^{4}

This is question of undamped free vibration of sinle degree of freedom system. For the above figure it is given that natural frequency and natural period are w rad/s and T s, respectively. The equation for soof system is
\omega =\sqrt{\frac{k}{m}}, T=\frac{2\pi}{\omega }
For the modified system, stiffness of the spring is doubled and the mass is halved.
Let us assume k'=2k and m'=m/2, here k' and m' represent the stiffness and mass of the modified system respectively.
\omega '=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{\left ( \frac{m}{2} \right )}}=\sqrt{\frac{4k}{m}}=2\sqrt{\frac{k}{m}}=2\omega \;\;\; \left ( \because \omega = \sqrt{\frac{k}{m}}\right )
T'=\frac{2\pi}{\omega '}=\frac{2 \pi}{2\omega }=\frac{1}{2}\left ( \frac{2 \pi}{\omega } \right )=\frac{T}{2} \;\;\left ( T=\frac{2 \pi}{\omega } \right )
\omega ' and T' are natural frequency and natural period of the modified system.
hence, the natural frequency of the modified system is doubled and natural period is halved compared to the original given system. Hence option (A) is correct.

\begin{aligned} \Sigma Y&=0 \Rightarrow \\ N_{2} \sin 60^{\circ}-0.2 N_{2} \sin &30^{\circ}-Q=0 \\ Q&=0.766 \mathrm{~N}_{2} \end{aligned}

\begin{aligned} \Sigma X &=0 \\ \Rightarrow \qquad 0.2 N_{2} \cos 30^{\circ}+N_{2} \cos 60^{\circ}-P &=0\\ P &=0.67 N_{2} \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad P &=0.67 \times \frac{Q}{0.766}\\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad P&=0.875 \mathrm{Q} \simeq 0.9 Q \\ P&=\alpha Q \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \alpha&=0.9 \end{aligned}

\begin{aligned} m\ddot{z}+kz&=0 \\ \ddot{z}+\frac{k}{m}z&=0 \\ \text{Comparing with} \\ \ddot{z}+\omega _n^2 z&=0 \\ \text{We get}\;\; \omega _n&=\sqrt{\frac{k}{m}} \end{aligned}

\begin{aligned} 12.5^{2} &=x^{2}+10^{2} \\ \Rightarrow \; \; x &=7.5\; \text{m} \end{aligned}
Bending moment at S=0 {consider the left part}
\begin{aligned} H_{P}\times 7.5 &=75\times 10 \\ H_{P} &=100\; \text{kN} \end{aligned}
Tension in cable
\begin{aligned} &=\sqrt{H_{P}^{2}+V_{P}^{2}}=\sqrt{75^{2}+100^{2}} \\ &=125\; \text{kN} \end{aligned}