The periodic time of one oscillation for a simple pendulum is:

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ISRO VSSC Technical Assistant Mechanical 7 April 2012 Official Paper
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  1. \(2\pi \sqrt {\frac{\gamma}{l}}\)
  2. \(\frac{1}{{2\pi }}\sqrt {\frac{\gamma}{l}}\)
  3. \(2\pi \sqrt {\frac{l}{g}}\)
  4. \(\frac{1}{{2\pi }}\sqrt {\frac{l}{g}}\)

Answer (Detailed Solution Below)

Option 3 : \(2\pi \sqrt {\frac{l}{g}}\)
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Detailed Solution

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Explanation:

F2 J.K 1.6.20 Pallavi D3

Here, we have for the ball to be in equilibrium

Tension in cord = mg cos θ

For small angle θ,

cos θ  = 1

So, F = mg sin θ

for the object in SHM

gθ = ω2

x = lθ

g = ω2

\(g\theta = {\left( {\frac{{2\pi }}{T}} \right)^2}l\theta\)

\(T = 2π{ \sqrt{l \over g}}\)

Time Period of Simple Pendulum

  • The time for completing one cycle is called a time period of a pendulum.
  • The time period of the pendulum watch depends upon the length of the pendulum and independent of the mass of the bob.
  •  \(T = 2π{ \sqrt{l \over g}}\) , where T = time period  L is the length of the pendulum and g is an acceleration of gravity.
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