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

Answer: 4

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