The shown frame is made from weightless rods of length 𝑙 and 2𝑙. ( left Fig.). All the joints have frictionless hinges. A massless spring of spring constant K is inserted into the middle link, after which the structure acquires the shape shown in left figure with a known angle $α .$ Then a load of mass m was suspended from below, and the angle became equal to $β$ ( right figure)If time period of oscillation when the load performs very small oscillations vertically is $2 π \sqrt{\frac{P m}{k}}$ , then find the value of P.

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Detailed Solution

Find the initial length of the spring $l_{0} = 2 l \sin \frac{α}{2}$ and the final one $l_{1} = 2 l \sin \frac{β}{2}$
Spring extension under load $Δ l = 2 l (\sin \frac{β}{2} - \sin \frac{α}{2}) T d x = K (Δ l + \frac{d x}{3})$

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To find tension To in equilibrium we use principal of virtual work. Let block move down by dx, the spring extends by dx/3. Hence $T_{o} d x = K (Δ l \frac{d x}{3}) o r T_{o} = K (\frac{Δ l}{3}) = m g$
If the block while oscillating moves down by small distance x $m a = T - m g = \frac{K (Δ l + x / 3)}{3} - m g$ or $a = - \frac{K x}{9}$