The natural frequency of a simple pendulum of length 1.0 m on earth's surface is :
[Take g = 9.8 m/s2]
0.5 Hz
Reason — Given,
length = 1.0 m
g = 9.8 m/s2
We know,
frequency (f) = 12πgL\dfrac{1}{2π} \sqrt\dfrac{g}{L}2π1Lg
Substituting we get,
frequency (f) = 12π9.81\dfrac{1}{2π} \sqrt\dfrac{9.8}{1}2π119.8
frequency (f) = 12π×3.13\dfrac{1}{2π} \times 3.132π1×3.13 = 0.49 ≈ 0.5 Hz
