1.1. To establish which factors influence the To establish which factors influence the period of a pendulumperiod of a pendulum

2.2. To understand how the period of a simple To understand how the period of a simple pendulum can be used to establish a value pendulum can be used to establish a value for g experimentallyfor g experimentally

Book Reference : Pages 42Book Reference : Pages 42

During the practical last lesson, we investigated During the practical last lesson, we investigated three factors which may have an effect upon the three factors which may have an effect upon the period of a simple pendulum:period of a simple pendulum:

1.1. Mass of the bobMass of the bob

2.2. Length of the stringLength of the string

3.3. Initial displacement (or amplitude)Initial displacement (or amplitude)

Which factors did you find affected the period?Which factors did you find affected the period?

[VPL : SHM][VPL : SHM]

L

Consider a simple Consider a simple pendulum with a bob pendulum with a bob of mass of mass mm suspended suspended by a thread of length by a thread of length LL..

Which has a Which has a displacement displacement ss from from the equilibrium the equilibrium position giving an position giving an angle angle to the vertical to the vertical

s

The weight of the bob (mg) has the following The weight of the bob (mg) has the following components:components:

Perpendicular to motion :Perpendicular to motion : mg cos mg cos Parallel to motion :Parallel to motion : mg sin mg sin

The restoring force The restoring force FF causing the SHM will be in causing the SHM will be in the opposite direction:the opposite direction:

F = -mg sin F = -mg sin (Using F=ma for a)(Using F=ma for a)

a = F/m = (-mg sin a = F/m = (-mg sin ) / m) / m

a = -g sin a = -g sin

For small values of For small values of , (< 10°), sin , (< 10°), sin = s/L = s/L

a = -g s/La = -g s/L

and we know that:and we know that:

Acceleration = - (2Acceleration = - (2f)f)2 2 x displacement x displacementa = - (2a = - (2f)f)2 2 ss

a = -g s/L = - (2a = -g s/L = - (2f)f)2 2 ss (remove s and -)(remove s and -)

g /L = (2g /L = (2f)f)22

f = f = (g /L) / 2(g /L) / 2 (since T = 1/f)(since T = 1/f)

T = 2T = 2 (L /g)(L /g)

From the equation we can see that:From the equation we can see that:

1.1. The period is independent of massThe period is independent of mass2.2. The period is independent of initial The period is independent of initial

displacement (amplitude)displacement (amplitude)3.3. The period is dependent upon the length of The period is dependent upon the length of

stringstring

More specifically TMore specifically T22 is proportional to L & and a is proportional to L & and a graph of Tgraph of T22 against L will have a gradient of 4 against L will have a gradient of 422/g/g

This can be used to establish a value for gThis can be used to establish a value for g

As the bob passes through the equilibrium As the bob passes through the equilibrium position....position....

The tension acts directly upwards and provides a The tension acts directly upwards and provides a centripetal forcecentripetal force

Tension – mg = mvTension – mg = mv22/L/L

This work was first carried out in 1581 by Galileo This work was first carried out in 1581 by Galileo when he observed lamps swinging backwards and when he observed lamps swinging backwards and forwards in the cathedral at Pisaforwards in the cathedral at Pisa

Since clocks had not yet Since clocks had not yet been invented he used been invented he used his own pulse for timing!his own pulse for timing!

Having conducted the experiment how could it be Having conducted the experiment how could it be improved?improved?

Why are the pendulum bobs in clocks not spherical? Why are the pendulum bobs in clocks not spherical?

Calculate the time period of a simple pendulum Calculate the time period of a simple pendulum with lengths with lengths

a.a. 1.0m1.0m

b.b. 0.25m0.25m

Take g to be 9.81 m/sTake g to be 9.81 m/s22

Now calculate the period for the 1.0m pendulum Now calculate the period for the 1.0m pendulum on the moon where gravity is around 1/6on the moon where gravity is around 1/6thth of of that on Earth (g=1.6 m/sthat on Earth (g=1.6 m/s22))