Answer:
the period of the physical pendulum is 0.498 s
Explanation:
Given the data in the question;
[tex]T_{simple[/tex] = 0.61 s
we know that, the relationship between T and angular frequency is;
T = 2π/ω ---------- let this be equation 1
Also, the angular frequency of physical pendulum is;
ω = √(mgL / [tex]I[/tex] ) ------ let this equation 2
where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and [tex]I[/tex] is moment of inertia of rod.
Now, moment of inertia of thin uniform rod D is;
[tex]I[/tex] = [tex]\frac{1}{3}[/tex]mD²
since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.
we substitute equation 2 into equation 1
we have;
T = 2π/ω OR T = 2π/√(mgL/[tex]I[/tex]) OR T = 2π√([tex]I[/tex]/mgL)
so we can use [tex]I[/tex] = [tex]\frac{1}{3}[/tex]mD² for moment of inertia of the rod
Since center of gravity of the uniform rod lies at the center of rod
so that L = [tex]\frac{1}{2}[/tex]D.
now, substituting these equations, the period becomes;
T = 2π/√([tex]I[/tex]/mgL) OR T = [tex]2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }[/tex] OR T = 2π√(2D/3g ) ----- equation 3
length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),
we have;
ω[tex]_{simple[/tex] = 2π/[tex]T_{simple[/tex] OR ω[tex]_{simple[/tex] = √(g/D) OR ω[tex]_{simple[/tex] = 2π√( D/g )
so we simple solve for D/g and insert into equation 3
so we have;
T = √(2/3) × [tex]T_{simple[/tex]
we substitute in value of [tex]T_{simple[/tex]
T = √(2/3) × 0.61 s
T = 0.498 s
Therefore, the period of the physical pendulum is 0.498 s
