What's your question? is it seprete questions
Answer:
What Electricity questions?
Explanation:
C. Maintain correct Posture
Answer:
e=mc2 made to relate mass with energy . bcoz energy can neither b created nor b destroyed
The equation of motion of a pendulum is:
![\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%7Bd%7D%5E2%5Ctheta%7D%7B%5Ctextrm%7Bd%7Dt%5E2%7D%20%3D%20-%5Cdfrac%7Bg%7D%7B%5Cell%7D%5Csin%5Ctheta%2C)
where
it its length and
is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles (
), we can use:
![\sin\theta \simeq \theta.](https://tex.z-dn.net/?f=%5Csin%5Ctheta%20%5Csimeq%20%5Ctheta.)
Additionally, let us define:
![\omega^2\equiv\dfrac{g}{\ell}.](https://tex.z-dn.net/?f=%5Comega%5E2%5Cequiv%5Cdfrac%7Bg%7D%7B%5Cell%7D.)
We can now write:
![\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctextrm%7Bd%7D%5E2%5Ctheta%7D%7B%5Ctextrm%7Bd%7Dt%5E2%7D%20%3D%20-%5Comega%5E2%5Ctheta.)
The solution to this differential equation is:
![\theta(t) = A\sin(\omega t + \phi),](https://tex.z-dn.net/?f=%5Ctheta%28t%29%20%3D%20A%5Csin%28%5Comega%20t%20%2B%20%5Cphi%29%2C)
where
and
are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:
![T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B2%5Cpi%7D%7B%5Comega%7D%20%3D%202%5Cpi%5Csqrt%7B%5Cdfrac%7Bg%7D%7B%5Cell%7D%7D.)
This justifies that the period depends only on the pendulum's length.