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Physics

The natural frequency of a simple pendulum of length 1.0 m on earth's surface is :

[Take g = 9.8 m/s2]

  1. 0.5 Hz
  2. 100 Hz
  3. 50 Hz
  4. 5 Hz

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Answer

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}

Substituting we get,

frequency (f) = 12π9.81\dfrac{1}{2π} \sqrt\dfrac{9.8}{1}

frequency (f) = 12π×3.13\dfrac{1}{2π} \times 3.13 = 0.49 ≈ 0.5 Hz

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