When permitted to oscillate, the time period is found to be $2.2 sec$. Find the dynamically equivalent system of 2 masses one of which located at the small end centre.

Solution:

tp = 2.2 sec.

$tp = 2\pi \sqrt{ \frac{k^2 + (0.74)^2}{9.81 \times 0.740}}$

$2.2 = 2\pi \sqrt{ \frac{k^2 + (.074)^2}{9.81 \times 0.741}}$

$K = 0.58 m$

$k^2 = l_1 \ l _2$

$(0.58)^2 = (0.700) \times l_2$

$l_2 = 0.489 \ m$

$m_1 = \frac{ml_2}{l_1 + l_2}$

$= \frac{42 \times 0.489}{0.700 + 0.489}$

$m_1 = 17.28 \ kg$

$m_2 = m - m_1$

$= 42 – 17.28$

$m_2 = 24.72 \ kg$