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1.To establish which factors influence the period of a pendulum 2.To understand how the period of a simple pendulum can be used to establish a value for g experimentally Book Reference : Pages 42

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During the earlier practical, we investigated 3 factors which may have an effect upon the period of a simple pendulum: 1.Mass of the bob 2.Length of the string 3.Initial displacement (or amplitude) Which factors did you find affected the period? [VPL : SHM]

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L Consider a simple pendulum with a bob of mass m suspended by a thread of length L. Which has a displacement s from the equilibrium position giving an angle to the vertical s

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The weight of the bob (mg) has the following components: Perpendicular to motion :mg cos Parallel to motion :mg sin The restoring force F causing the SHM will be in the opposite direction: F = -mg sin (Using F=ma for a)

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a = F/m = (-mg sin ) / m a = -g sin For small values of , (< 10°), sin = s/L a = -g s/L and we know that: Acceleration = - (2 f) 2 x displacement a = - (2 f) 2 s

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a = -g s/L = - (2 f) 2 s(remove s and -) g /L = (2 f) 2 f = (g /L) / 2 (since T = 1/f) T = 2 (L /g)

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From the equation we can see that: 1.The period is independent of mass 2.The period is independent of initial displacement (amplitude) 3.The period is dependent only upon the length of string and g. More specifically T 2 is proportional to L & and a graph of T 2 against L will have a gradient of 4 2 /g This can be used to establish a value for g experimentally

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As the bob passes through the equilibrium position.... The tension acts directly upwards and provides a centripetal force Tension – mg = mv 2 /L

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This work was first carried out in 1581 by Galileo when he observed lamps swinging backwards and forwards in the cathedral at Pisa Since clocks had not yet been invented he used his own pulse for timing!

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Having conducted the experiment how could it be improved? Why are the pendulum bobs in clocks not spherical?

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Calculate the time period of a simple pendulum with lengths a.1.0m [2.0s] b.0.25m [1.0s] Take g to be 9.81 m s -2 Now calculate the period for the 1.0m pendulum on the moon where gravity is around 1/6 th of that on Earth (g=1.6 m s -2 ) [5.0s]

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Whenever the force acting on an object is: Whenever the force acting on an object is: 1. Proportional to the displacement 2. In the opposite direction,

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