Question

Two identical, thin rods, each with mass m and length L, are joined at right angles to form an L-shaped object. This object is balanced on top of a sharp edge. If the L-shaped object is deflected slightly, it oscillates.
Find the frequency of oscillation.

Answer 1

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.

Answer 2

Moment of inertia for one rod is expressed as:

I(1)= I(end) + md^2 = (1\12)mL^2+m(L/2√2)^2=(5/24)mL^2

Therefore, for two rods:

I2 = 2I1 = (5/12)mL^2  


For the moment of inertia at the pivot point,

I=I2+2md^2=(5/12)mL^2+2(m(L/2√2)^2)=(2/3)mL^2 

Substituting the equations above to the equation for frequency:

f=(1/2π)√(2mgd/I)=(1/4π)√(6g/√2L)