Question

Prove(show) ''T=2π√(l/g)''​

Answer 1

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

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

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