Show that T =2/π√l/g is correct with the help of dimension
Answers
Answer:Forgive me if I assume you only know some good algebra and some basic physics, such as the equations of circular motion and SHM (simple harmonic motion.) With these constraints in mind I will try to show you how to prove the equation you are working on, based only on pure physics arguments, some simplifying assumptions, and very little mathematics. Once you get done with the explanations, it will then look easy!
But first, a couple of things from circular motion and SHM:
The speed of a particle in uniform circular motion around a circle of radius r is
v=ωrv=ωr
where ω=2π/Tω=2π/T. The acceleration of the particle is
a=ω2ra=ω2r
Now, let us see how we can apply all these in the case of the pendulum.
It is very crucial to understand that the equation you are trying to prove holds only for very small deviations of the bob (the mass) from its “middle” equilibrium position.
Now imagine you have the mass mm hanging vertically from some fixed point, by a piece of light and inextensible string of length LL. Give the mass a very small deviation from the equilibrium position. As the mass oscillates it is moving on an arc of a vertical circle of radius r=Lr=L, but the deviation must be small. The smaller the deviation the better, because otherwise even using calculus will not lead to the simple equation we want to prove. In fact, although the formal mathematical derivation requires a small deviation, it does not set a limit how small it must be. So we shall take it to be so small, that the centripetal force that holds the particle in the circular motion is the weight, mgmg, of the bob. That means the centripetal acceleration is gg. Therefore, according to the second of the above equations we can write
g=ω2Lg=ω2L
g=4π2T2Lg=4π2T2L
Solve the last equation for T to you get: T=2πLg−−√T=2πLg.
Hope this Helps You Bro!
Explanation: