Respuesta :
Answer:
Time period for Simple pendulum, [tex]T=2\pi\sqrt{\frac{l}{g}[/tex]
Explanation:
The Simple Pendulum
Consider a small bob of mass [tex]m[/tex] is tied to extensible string of length [tex]l[/tex] that is fixed to rigid support. The bob is oscillating in the plane about verticle.
Let [tex]\theta[/tex] is the angle made by string with vertical during oscillation.
Vertical component of the force on bob, [tex]F=-mg\sin\theta[/tex]
Negative sign shows that its opposing the motion of bob.
Taking [tex]\theta[/tex] as very small angle then, [tex]\sin\theta\sim\theta[/tex]
[tex]F=-mg\theta[/tex]
Let [tex]x[/tex] is the displacement made by bob from its mean position ,
then, [tex]\theta=\frac{x}{l}[/tex]
so, [tex]F=-mg\frac{x}{l}[/tex] ........(1)
Since, pendulum is in hormonic motion,
as we know, [tex]F=-kx[/tex]
where [tex]k[/tex] is the constant and [tex]k=m\omega^{2}[/tex]
[tex]F=-m\omega^2x[/tex] .........(2)
From equation (1) and (2)
[tex]-m\omega^2x=-mg\frac{x}{l}[/tex]
[tex]\omega=\sqrt{\frac{g}{l}}[/tex]
Since, [tex]\omega=\frac{2\pi}{T}[/tex]
[tex]\frac{2\pi}{T}=\sqrt{\frac{g}{l}[/tex]
[tex]T=2\pi\sqrt{\frac{l}{g}}[/tex]
