Question

Four weightless rods of length $l$ each are connected by hinged joints and form a rhombus as shown in figure. Hinge A is fixed and load of mass m is suspended to hinge C. Hinges B and D are connected by weightless spring of length $1.5l$ in undeformed state. At equilibrium $\theta ={30}^o$, then the time period of SHM of block if displaced slightly is $T=2\pi \sqrt{\frac{nm}{k}}$, here $n$ is [Write upto two digits after the decimal.]

Answer 1

$2T\sin \theta =F_R=k[1.5l-2l\sin \theta ]$ --(1)
and $F=2T\cos \theta -mg$
or
$F=\frac{k\cos \theta }{\sin \theta }[1.5l-2l\sin \theta ]-mg$
By differentiating with $\theta$, we get
$dF=-5kld\theta$ $[for\theta ={30}^o]-\left(2\right)$
We know that,
$y=2l\cos \theta ⟹dy=-2l\sin \theta d\theta$ --(3)
From (2) and (3),
$\therefore dF=5kdy$
$a=\frac{dF}{m}=\frac{5k}{m}dy$ in the upward direction, so it is restoring in nature.
$⟹\omega =\sqrt{\frac{5k}{m}}$
$⟹T=2\pi \sqrt{\frac{m}{5k}}$