Answer:
T = 1.54 s
Explanation:
given,
length of rod = 0.928 m
[tex]T = 2\pi \sqrt{\dfrac{I}{MgD}}[/tex]
D is distance from pivot to CM
[tex]D = \dfrac{L}{2}-0.043[/tex]
[tex]D = \dfrac{0.928}{2}-0.043[/tex]
D = 0.421 m
[tex]T = 2\pi \sqrt{\dfrac{I}{MgD}}[/tex]
[tex]T = 2\pi \sqrt{\dfrac{\dfrac{1}{12}ML^2+Mx^2}{MgD}}[/tex]
[tex]T = 2\pi \sqrt{\dfrac{\dfrac{1}{12}L^2+x^2}{gD}}[/tex]
[tex]T = 2\pi \sqrt{\dfrac{\dfrac{1}{12}\times 0.928^2+0.421^2}{9.8 \times 0.421}}[/tex]
T = 1.54 s
The period of oscillation will be "1.54 s".
According to the question,
Length of rod = 0.928 m
The distance from pivot to CM will be:
→ [tex]D = \frac{L}{2}-0.043[/tex]
[tex]= \frac{0.928}{2} - 0.043[/tex]
[tex]= 0.421 \ m[/tex]
hence,
The period of oscillation be:
→ [tex]T = 2 \pi \sqrt{\frac{I}{MgD} }[/tex]
[tex]= 2 \pi \sqrt{\frac{\frac{1}{2}ML^2+Mx^2}{MgD} }[/tex]
[tex]= 2 \pi \sqrt{\frac{\frac{1}{12}\times 0.928^2+0.421^2 }{9.8\times 0.421} }[/tex]
[tex]= 1.54 \ s[/tex]
Thus the above answer is correct.
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