time period of simple pendulum of length L is T1 and time period of uniform rod of the same length L is pivoted about one end and oscillating in a vertical plane is T2 .amplitude in both the oscillations in both the case is small.find T1/T2
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For T1 we have to use formula
T 1 = √L/g as it is a simple pendulum.
T 2 = √I/mgy as it is compound pendulum.
T 1 = √L/g
T 2 = √mL^2/3mgL
= √2L/3
T 1 / T 2 = √L/g ÷ √2L/3
= √3/2.
Hope it helps!
T 1 = √L/g as it is a simple pendulum.
T 2 = √I/mgy as it is compound pendulum.
T 1 = √L/g
T 2 = √mL^2/3mgL
= √2L/3
T 1 / T 2 = √L/g ÷ √2L/3
= √3/2.
Hope it helps!
Jhani:
Pls mark it best if understood.
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6
The value of T1T2 is √(3/2)
Time period of length = L = T1 (Given)
Thus, T1 = √ (L/g) as it is a simple pendulum.
Time period in vertical plane = T2 (Given)
T2 = √ (I)/ (mgy) as it is compound pendulum.
Therefore,
T1 = √ (L/g)
T2 = √ (mL²/ (3mgL)
= √ (2L/3)
(T1 / T2)= √ (L/g) / √ (2L/3)
= √(3/2)
Thus, the value of T1T2 = √(3/2)
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