The Reflection Of The Definition Of A Simple Pendulum

A pendulum follows a very simple type of oscillation called “isochronous”, which means it always takes the same time. Therefore these are oscillations that always repeat in the same time period maintaining the same this constant time even with a change in amplitude. A simple pendulum follows this pattern and it is why it has been such an important element in clocks. This can also be applied in different gravitational field strengths as the motion of the pendulum will always take the same time for one oscillation. Although there is a limit to the angle of a simple pendulum on the earth as if the amplitude is too much the pendulum will be slowed down by air friction.
When a pendulum bob is displaced from its equilibrium position, hanging vertically, …show more content…
This can also be described using frequency per oscillation, which measures how many oscillations it does per second. Frequency is inverse of the period, f = 1/T. Also the period is inverse of the frequency, T = 1/f. We know that the horizontal component of the force is Tsinθ = -ma and the vertical component of the force is Tcosθ = mg. Therefore we can say that:
Tsinθ/Tcosθ= (-ma)/mg tanθ= (-a)/g
In our investigation we are going to keep the angle θ of the pendulum very small therefore we can say that the length L is going to be approximately the same for T. Also tanθ will approximately equal to sinθ. sinθ=(-a)/g x/l= (-a)/g ∴a=(-x)/l g

Using the laws of circular motion we can calculate the time period using the values for a we found above. w=2πf a=-(〖w)〗^2 x
Where w is the angular speed and f is the frequency. Using the equation above for a we can set it equal to the one we found previously to find T using f from angular speed.
(-x)/g=-(〖2πf)〗^2 x g/l=(〖2πf)〗^2 √(g/l)=2πf f=√(g/l)/2πf T=2π√(l/g)
We can also solve for g to find the acceleration due to gravity from our result by rearranging the equation.
g=(4π^2