how does the length of the pendulum and the mass of the bob affect the period of the pendulum?? whats the logic behind??

how does the length of the pendulum and the mass of the bob affect the period of the pendulum?? whats the logic behind??
(someone correct me if I'm wrong)
For length :
As you increase the length of the string, the distance the bob has to travel to complete an oscillation increases and thus the period increases.
For mass :
Since no energy is lost, kinetic energy = potential energy
1/2 mv^2 = mgh > v = square root (2gh)
So mass actually doesn't affect the speed and thus doesn't affect the period.
The simplest way to derive the equations of motion for a mathematical pendulum is by considering the energy equation: mgh + ½mv^2 = E_0 at all times. Using the constraints imposed by the pendulum, we may rewrite this as mgl(1cos(\theta)) + ½ml^2*\thetadot^2 = E_0
Since energy is conserved, E_0 is constant with time. By taking the differential with respect to time on both sides of the equation we arrive at mgl*sin(\theta)*\thetadot + ml^2*\thetadoubledot*\thetadot = 0
Since \thetadot =/= 0 (except at the ends of the motion, which are trivially easy to solve anyway), we may reduce this to \thetadoubledot =  g/l*sin(\theta). The Taylor series of sin(x) is x  1/6x^3 + higher orders.
So as long as the angle is so small that \theta >> 1/6\theta^3 we may substitute \theta for sin(\theta), and arrive at the differential equation \thetadoubledot = g/l*\theta This equation can be solved and the solution is \theta = A*cos(sqrt(g/l)*t + \phi)
A and \phi are determined by the constraints of the problem. This is a onedimensional harmonic motion with a period of 2\pi*sqrt(l/g)
Does that answer your question?
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