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Q. Three identical ideal springs, each of force constant $k$ are joined to three identical balls (each of mass $m$ ) as shown in the figure. $O$ is centroid of the triangle. Initially, each of the springs is in its natural length. Now all the three balls are simultaneously given small displacements of equal magnitude along the directions shown in the figure. The oscillation frequency of balls is $\frac{1}{2 \pi} \sqrt{\frac{\beta k}{2 m}}$. The value of $\beta$ is ____Physics Question Image

Oscillations

Solution:

In the figure, $\theta=30^{\circ}$.
When displacement is each ball $x$,
the extension in each spring is $x^{\prime}=x \cos \theta=x \frac{\sqrt{3}}{2}$
Each spring is connected to two balls.
So, net extension of each spring $=2 x^{\prime}=\sqrt{3} x$
Force along the direction of displacement on each ball is
$F=k \sqrt{3} x \cos 30^{\circ}+k \sqrt{3} x \cos 30^{\circ}=3 k x$
$F=-3 k x$ [negative because restoring in nature]
$a=\frac{F}{m}=-\frac{3 k}{m} x$
$ \Rightarrow \omega=\sqrt{\frac{3 k}{m}}$
$f=\frac{2}{2 \pi} \sqrt{\frac{3 k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{2 m}}$
$ \Rightarrow \beta=6$