EG260 Solutions-example_1

# EG260 Solutions-example_1 - EG-260 Examples Sheet 1 1 This...

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Unformatted text preview: EG-260 Examples Sheet 1 1. This is a rotational problem, so we express Newton's Law in the form & C = I& where C is the applied moment and I is the moment of inertia. In the pendulum case C = -mgl sin I = ml 2 So that & - mgl sin = ml 2& For small angles we write sin = then & ml 2& + mgl = 0 & g & + = 0 l This is our standard form n2 = g l In a clock, T= 2 =2 n l =1 g g 9.81 = = 0.994 m 2 9.87 l = @Avril Slone 2007 EG-260 2. n = k 4 = =2 m 1 The general solution to the equation of motion we can express as x(t ) = A sin 2t + B cos 2t At t = 0 , x(0) = B = 1 & x(0) = 2 A = 0 Hence x(t ) = cos 2t which is readily plotted using MATLAB. 3. n = k 400000 = = 20 m 1000 We are not given any initial conditions, so the form of the deflection will be x(t ) = A cos 20t where A = 0.1 (10cm) The new natural frequency is 400000 = 17.54 rad/s 1300 The velocity amplitude is 0.1 17.54 = 1.754 ms -1 The acceleration amplitude is 0.1 17.54 2 = 30 .769 ms -2 4. SHM - Acceleration amplitude A n = 10 ms -2 where n = 2 f = 2 8 rad s -1 10 SHM maximum displacement is A = 2 = 0.00396 m = 0.396 cm n 2 @Avril Slone 2007 ...
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## This note was uploaded on 05/18/2010 for the course ENGINEERIN EG 260 taught by Professor Stone during the Spring '10 term at Swansea UK.

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EG260 Solutions-example_1 - EG-260 Examples Sheet 1 1 This...

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