Calculate the length of a second pendulum having a time period of 2s at a plate where g = 9.8 m s⁻². (L = 0.992 m)
Given:
The time period of the second pendulum = T = 2 s
Value of acceleration due to gravity = g = 9.8 m s⁻²
To Find:
Solution:
The formula for the Time Period of the pendulum is
T = 2π `\sqrt frac{L}{g}`
To separate the L from this formula we will take square on both sides and by simplifying we get
L = `\frac {T^2 g}{(2π)^2}`
by putting values
L = `\frac {(2s)^2 x 9.8 m s⁻²}{(2 x 3.1416)^2}`
L = `\frac {4 s² x 9.8 m s⁻²}{(6.283185)^2}`
L = `\frac {39.2 m}{39.478}`
L = 0.9929 m ------------Ans.
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© 2022-Onwards by Academic Skills and Knowledge (ASK)
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