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IRINA_888 [86]
3 years ago
5

Prove(show) ''T=2π√(l/g)''​

Physics
1 answer:
Nonamiya [84]3 years ago
6 0

Answer:

Time period for Simple pendulum, T=2\pi\sqrt{\frac{l}{g}

Explanation:

The Simple Pendulum

Consider a small bob of mass m is tied to extensible string of length l that is fixed to rigid support. The bob is oscillating in the plane about verticle.

       Let \theta is the angle made by string with vertical  during oscillation.

Vertical component of the force on bob, F=-mg\sin\theta

Negative sign shows that its opposing the motion of bob.

Taking \theta as very small angle then, \sin\theta\sim\theta

F=-mg\theta    

Let x is the displacement made by bob from its mean position ,

then, \theta=\frac{x}{l}

so, F=-mg\frac{x}{l}                ........(1)

Since, pendulum is in hormonic motion,

as we know, F=-kx

where k is the constant and k=m\omega^{2}

F=-m\omega^2x                   .........(2)

From equation (1) and (2)

-m\omega^2x=-mg\frac{x}{l}

\omega=\sqrt{\frac{g}{l}}

Since, \omega=\frac{2\pi}{T}

\frac{2\pi}{T}=\sqrt{\frac{g}{l}

T=2\pi\sqrt{\frac{l}{g}}

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