Question

Prove(show) ''T=2π√(l/g)''​

Answer 1

Time period for Simple pendulum,
`T=2\pi\sqrt{(l)/(g)`

Step-by-step 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=(x)/(l)`

so,
`F=-mg(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(x)/(l)`

`\omega=\sqrt{(g)/(l)}`

Since,
`\omega=(2\pi)/(T)`

`(2\pi)/(T)=\sqrt{(g)/(l)`

`T=2\pi\sqrt{(l)/(g)}`