4.6a Factors affecting the period of a pendulum We have looked at pendula oscillating in terms of their displacement, velocity and acceleration. We now need to investigate the factors that effect their period of oscillation. We should already know from initial investigations that length is the only factor that effects the period however we can check this quickly. Adjust the length, amplitude or mass of the pendulum through a range, repeat the readings enough times to eliminate errors. Record the period of oscillations. Plot an appropriate graph so the relationship between them is obvious and the gradient can be used to calculate acceleration due to gravity. Try a few common functions, if you can’t find the answer by trial and error look it up!

Period of a Pendulum This derivation should be understood fully rather than learned verbatum!  l x Consider a pendulum of length “ l ” and mass “ m ” suspended by a light thread at an angle to the vertical. The pendulum is moved a small amount to the side so that the bob can be considered to have moved no distance vertically. The angle of inclination is small enough such that cos() = 1.

Fsin() = ma from Newton’s second law If we consider the forces acting on the bob we have a situation looking like this. Assuming the bob moves very small distances vertically, there can be considered to be no vertical acceleration, therefore the vertical component of the tension balances the weight. So: Fcos() + mg = 0 and since cos() = 1 for small angles we get F = - mg  mg F Fsin() Fcos() As the horizontal motion is the only one where we can consider there to be an acceleration we have the situation: Fsin() = ma from Newton’s second law

Fsin() = ma from Newton’s second law and So using the equation: Fsin() = ma from Newton’s second law and F = - mg We get -mgsin() = ma sin() = -a/g  l x From this diagram we can see that sin() = x / l Therefore… -a/g = x / l a = -x g/ l If we compare this to the equation for SHM a = - 2x = - (2f)2x

  2  T = l g So if a = - (2f)2x = -xg/l We can see that (2f)2 = g/l This gives us the equation: 2  l F = 1 g  2  T = l  g Now investigate the theory to prove it to be correct!

Try plotting the length of the pendulum against period of oscillation. If it is a curve plot period against log(length) this will give you a straight line, the gradient of which is the “power” to which the length is e.g. Grad = 2 means in the equation length is squared etc.