Question

Two identical small discs each of mass $m$ placed on a frictionless horizontal floor are connected with the help of a light spring of force constant $k$. The discs are also connected with two light rods each of length $2 \sqrt{2} \,m$ that are pivoted to a nail driven into the floor as shown in the figure by a top view of the situation. If period of small oscillations of the system is $2 \pi \sqrt{(m / k)}$, find relaxed length (in meters) of the spring.

Answer 1

$2 k x-T \cos \theta=m a$
Applying torque equation about point $A$.
$2 k x \cdot L \sin \theta=M L^{2} \cdot x$
Applying constraint equation
$L \alpha \sin \theta=a$
$2 k x \cdot \sin ^{2} \theta=M a$
$2 k x \sin ^{2} \theta=m \cdot\left(\omega^{2} x\right)\,\, (\because a = \omega ^{2}x)$
$\Rightarrow 2 k \sin ^{2} \theta=m\left(\frac{k}{m}\right)$
$\Rightarrow 2 \sin ^{2} \theta=1$
$\sin ^{2} \theta=\frac{1}{2}$
$\sin ^{2} \theta=\frac{1}{\sqrt{2}} $
$\Rightarrow \theta=45^{\circ}$
$\Rightarrow$ Natural length of spring $=2 L \cos \theta$
$=2 \times(2 \sqrt{2}) \times \frac{1}{\sqrt{2}}=4\, m$