What is the natural frequency for a child's swing with 2.00 m long chains?

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The discussion centers around calculating the natural frequency of a child's swing with 2.00 m long chains. The formula for the period of a pendulum is provided as T = 2π√(l/g), where l is the length of the swing and g is the acceleration due to gravity. The largest amplitude occurs at the natural frequency, which is derived from this period. Clarifications are sought regarding the transformation of terms in the frequency formula and the application of the small angle approximation. Understanding these concepts is crucial for determining the optimal frequency for pushing the swing.
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Please help me with this, it's one of the 'basic' problems and I can't do it at all. "The chains suspending a child's swing are 2.00 m long. At what frequency should a big brother push to make the child swing with largest amplitude?"

Okay, so... the largest amplitude comes with a natural frequency...
f1 = (1/2L)*(T/mu)^(1/2)

I'm trying to figgure out why the rooted part gets to (g/L)^1/2... if T is mg and mu is m/L, wouldn't it be (gL)^1/2? And how does 1/2L turn into 1/2pi? Thanks for any help you can offer.
 
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You are using a formula for transverse waves.

With the small angle approximation, the period of a pendulum:
T = 2\pi \sqrt{\frac{l}{g}}
 
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