The frequency of oscillation of the given L-shaped object made of two rods is [tex]\boxed{\frac{1}{{2\pi }}\sqrt{\frac{{3g}}{{4\sqrt2L}}}}[/tex] .
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 [tex]45^\circ[/tex] with the sharp edge. When there is a very small deflection in the rod, it will start performing oscillations with a small angle [tex]\theta[/tex] . Then the time period of oscillation of this rod about its center of gravity is given as:
[tex]T=2\pi \sqrt {\frac{I}{{mgd}}}[/tex] …… (1)
Here, [tex]I[/tex] is the moment of inertia of the rod, [tex]d[/tex] is the distance of the center of gravity from the fixed point and [tex]T[/tex] 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:
[tex]I=\frac{1}{3}M{L^2}[/tex]
For the above mentioned case, the moment of inertia of the two rods about the fixed sharp point is:
[tex]\begin{aligned}I&=\frac{1}{3}m{L^2}+\frac{1}{3}m{L^2}\\&=\frac{2}{3}m{L^2}\\\end{aligned}[/tex]
The distance of the center of gravity from the fixed point is:
[tex]d=\frac{L}{{2\sqrt 2 }}[/tex]
Substitute the values in equation (1).
[tex]\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}[/tex]
The frequency of oscillation of the L-shaped object is:
[tex]\begin{aligned}f&=\frac{1}{T}\\&=\frac{1}{{2\pi}}\sqrt{\frac{{3g}}{{4\sqrt 2 L}}}\\\end{aligned}[/tex]
Thus, the frequency of oscillation of the given L-shaped object made of two rods is
[tex]\boxed{\frac{1}{{2\pi }}\sqrt{\frac{{3g}}{{4\sqrt 2 L}}}}[/tex] .
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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.
The amount of time in which an action or a state continues to exist. Due to the nature of the activity or situation under examination, it can be measured in seconds or millions of years. so, the calculted time is "[tex]\bold{2\pi \sqrt{\frac{\sqrt{2}L}{3g}}}[/tex]" .
The rod has a mass of m.
L is the length of a single rod.
If it's connecting at [tex]\bold{90^{\circ}}[/tex], as per the query,
Following that the combined system's moment of inertia.
[tex]\to \frac{mL^2}{3} + \frac{mL^2}{3}=\frac{2mL^2}{3}[/tex]
The pendulum's effective length:
[tex]\to L \cos 45^{\circ}= L \times \frac{1}{\sqrt{2}} =\frac{L}{\sqrt{2}}\\\\[/tex]
Calculating the time period:
[tex]T =2\pi \sqrt{\frac{I}{mgl}} \\\\[/tex]
[tex]= 2\pi \sqrt{\frac{2\times mL^2 \times \sqrt{2}}{3 \times 2 mgL}}\\\\[/tex]
[tex]= 2\pi \sqrt{\frac{L\sqrt{2}}{3 \times g}} \\\\= 2\pi \sqrt{\frac{\sqrt{2}L}{3g}}[/tex]
Find out more about the time period here:
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