The frequency of oscillation of the given L-shaped object made of two rods is \(\boxed{\frac{1}{{2\pi }}\sqrt{\frac{{3g}}{{4\sqrt2L}}}}\) .
Further Explanation:
The combination of the two rods perpendicular to each other and making a L-shaped object is as shown in figure attached below.
This L-shaped object rests on a sharp edge and therefore, each rod will make an angle of \(45^\circ\) with the sharp edge. When there is a very small deflection in the rod, it will start performing oscillations with a small angle \(\theta\) . Then the time period of oscillation of this rod about its center of gravity is given as:
\(T=2\pi \sqrt {\frac{I}{{mgd}}}\) …… (1)
Here, \(I\) is the moment of inertia of the rod, \(d\) is the distance of the center of gravity from the fixed point and \(T\) is the time period of oscillation of the L-shaped object.
The moment of inertia of the rod about its one end is given as:
\(I=\frac{1}{3}M{L^2}\)
For the above mentioned case, the moment of inertia of the two rods about the fixed sharp point is:
\(\begin{aligned}I&=\frac{1}{3}m{L^2}+\frac{1}{3}m{L^2}\\&=\frac{2}{3}m{L^2}\\\end{aligned}\)
The distance of the center of gravity from the fixed point is:
\(d=\frac{L}{{2\sqrt 2 }}\)
Substitute the values in equation (1).
\(\begin{aligned}T&=2\pi\sqrt{\frac{{\frac{2}{3}m{L^2}}}{{mg\left({\frac{L}{{2\sqrt 2 }}}\right)}}}\\&=2\pi\sqrt{\frac{{4\sqrt 2 L}}{{3g}}}\\\end{aligned}\)
The frequency of oscillation of the L-shaped object is:
\(\begin{aligned}f&=\frac{1}{T}\\&=\frac{1}{{2\pi}}\sqrt{\frac{{3g}}{{4\sqrt 2 L}}}\\\end{aligned}\)
Thus, the frequency of oscillation of the given L-shaped object made of two rods is
\(\boxed{\frac{1}{{2\pi }}\sqrt{\frac{{3g}}{{4\sqrt 2 L}}}}\) .
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Answer Details:
Grade: Senior School
Subject: Physics
Chapter: Simple Harmonic Motion
Keywords:
L-shaped object, two identical thin rods, balanced, a sharp edge, deflected slightly, moment of inertia, frequency of oscillation, center of gravity.
