Having done some work on determining pi by mathematical methods, I'm now going to use it in conjunction with some experimental work to determine the value of g, which is acceleration due to gravity. Any reference book will tell you the value of g is approximately 9.81 ms-2, but I'm going to do an experiment to show what it is. It's not a difficult experiment, and it doesn't require any specialised scientific equipment. To give you an idea, I did this using a toddler fireguard for my vertical surface, a piece of sewing cotton for my pendulum, and in the absence of any respectable small mass, used a small pine cone tied to the end of it. I also used a standard stopwatch on a digital watch (it's accurate to 1/100th of a second, although I'm not).

The relationship between a pendulum and g is described in the following limerick:

In order to improve the accuracy of my results, I counted the time taken for ten complete 'swings' or periods, and quoted this. I also repeated each ten-swing measurement three times, so that I could take an average and identify any anomalies. And somehow, saying that, I feel like I'm writing up a GCSE science coursework piece!

Here are my results...

I calculated the average value of 10T, and hence T and then T

However, I'm going at it in number-crunching form, using the formula above. My results for g are as follows:

So, not perfect, but given the nature of the experiment - me with a fireguard and a pine cone - it's not too bad at all, and I feel quite pleased at having worked out something so massively significant with such basic equipment, and I feel it proves that science isn't just for big-budget departments!

Next time, determining the distance to the moon using the same principle as for geostationary satellites (except that this one is a bit bigger, a bit further away and

The relationship between a pendulum and g is described in the following limerick:

*If a pendulum's swinging quite free*

Then it's always a marvel to me

That each tick plus each tock

Of the grandfather clock

Is 2 pi root L over gThen it's always a marvel to me

That each tick plus each tock

Of the grandfather clock

Is 2 pi root L over g

In order to improve the accuracy of my results, I counted the time taken for ten complete 'swings' or periods, and quoted this. I also repeated each ten-swing measurement three times, so that I could take an average and identify any anomalies. And somehow, saying that, I feel like I'm writing up a GCSE science coursework piece!

Here are my results...

length (l, in metres) | 10 swings (10T, seconds) | Run 2 | Run 3 |

0.162 | 8.45 | 8.31 | 8.35 |

0.237 | 10.09 | 9.90 | 10.04 |

0.321 | 11.63 | 11.67 | 11.66 |

0.344 | 12.09 | 11.90 | 12.17 |

0.410 | 12.98 | 12.95 | 13.00 |

0.475 | 14.23 | 14.13 | 14.03 |

I calculated the average value of 10T, and hence T and then T

^{2}, which I can use to determine g, with the following rearrangement:
An alternative, if I'd wanted to plot a graph of my data, is to determine g by finding the slope of the appropriate plot. Using the following rearrangement, it's possible to plot T

^{2}against l and have a slope of 4pi^{2}/gHowever, I'm going at it in number-crunching form, using the formula above. My results for g are as follows:

length (l, in metres) | g in ms^{-2} |

0.162 | 9.136 |

0.237 | 9.338 |

0.321 | 9.332 |

0.344 | 9.3436 |

0.410 | 9.612 |

0.475 | 9.395 |

So, not perfect, but given the nature of the experiment - me with a fireguard and a pine cone - it's not too bad at all, and I feel quite pleased at having worked out something so massively significant with such basic equipment, and I feel it proves that science isn't just for big-budget departments!

Next time, determining the distance to the moon using the same principle as for geostationary satellites (except that this one is a bit bigger, a bit further away and

*not*geostationary!).
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