**1 Answer**

written 3.9 years ago by |

**Time of oscillation when suspended from small end $= 175 sec$ and from big and $= 1.6 sec$. Determine m – I about the axis passing there mass centre (ii) The dynamically equivalent system of 2 mass one located at small and centre.**

**Solution:** $L = L_1 + L_2 = 800mm$

$tp_1 = 1.75$ sec/cycle.

$tp_1 = 2 \pi \sqrt{ \frac{k^2 + (L'_1)^2}{g.L'_1}}$

$1.75 = 2 \pi \sqrt{ \frac{k^2 (L_1 + 0.03)^2}{g (L_1 + 0.03)}}$

$k^2 = 0.701 L_1 + 0.0237 – L_1^2$

$tp_2 = 2\pi \sqrt{\frac{k^2+ (L'_2)^2}{gL'_2}}$

Where,

$L'_1 = L_1 + 0.03$

$L'_2 = L_2 + 0.04$

But $L_2 = 0.8 – L_1$

So $L'_2 = 0.8 – L_1 + 0.04$

$L'_2 = 0.84 – L_1$

{For big end}

$1.6 = 2 \pi \sqrt{ \frac{k^2+ (0.84 - L_1)^2}{9.81 (0.84 – L_1)}}$

$k^2 = 1.044 L_1 – 0.1717 – L_1^2$

$0.701 L_1 + 0.0237 – L_1^2 = 1.044 L_1 – 0.1717 – L_1^2$

$L'_1 = 0.569 \ m$

$L'_1 = 0.569 + 0.030$

$L'_1 = 0.599$

$L_2 = 0.800 – 0.569$

$L_2 = 0.2303$

Substituting value of $L_1$

$K = 0.31m$

$\therefore$ $I_G = mK^2$

$=40 (0.31)^2$

$= 43.9 \ kgm^2$

**2]**

$k^2 = l_1 . l_2$

$(0.314)^2 = 0.569 . l_2$

$l_2 = 0.173 \ m$

$m_2 = \frac{mL_1}{L_2} = \frac{40\times 0.5694}{0.569 + 0.173} = 30.69 \ kg$

$m_1 = \frac{mL_2}{L_1 + L_2} = 9.31kg$