To determine The displacement as a function of time. Explanation Formula used: The quadratic equation a x 2 + b x + c = 0 is calculated as x = − b ± b 2 − 4 a c 2 a . Theorem used: If r 1 = α + j β and r 1 = α − j β are the two complex roots of the auxiliary equation a r 2 + b r + c = 0 , then y = e α x ( c 1 cos β x + c 2 sin β x ) is the general solution of a y ″ + b y ′ + c y = 0 . Calculation: The simple harmonic motion equation is: D 2 x + 0.2 D x + 100 x = 0 The auxiliary equation is: m 2 + 0.2 m + 100 = 0 Comparing the above equation with a x 2 + b x + c = 0 . Here a = 1 , b = 0.2 , c = 100 . Substitute the value of a = 1 , b = 0.2 , c = 100 in the quadratic formula x = − b ± b 2 − 4 a c 2 a . m = − 0.2 ± 0.2 2 − 4 × 1 × 100 2 × 1 = − 0.2 ± 0.04 − 400 2 = − 0.1 ± 0.01 − 100 = − 0.1 ± 10 j So, the general solution to the above differential equation is x = e − 0.1 t ( c 1 sin 10 t + c 2 cos 10 t ) Substitute the initial conditions x = 4 when t = 0 in x = e − 0.1 t ( c 1 sin 10 t + c 2 cos 10 t ) . 4 = e − 0.1 × 0 ( c 1 sin 10 × 0 + c 2 cos 10 × 0 ) 4 = c 2 Differentiate the equation x = e − 0

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780134434636

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 31.10, Problem 3E

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The displacement as a function of time.

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Chapter 31 Solutions

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

Ch. 31.1 - Show that is a solution of . Is it the general...Ch. 31.1 - Prob. 1E Ch. 31.1 - In Exercises 1 and 2, show that the indicated...Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 4E Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 6E Ch. 31.1 - In Exercises 7–10, show that each function y =...Ch. 31.1 - Prob. 8E Ch. 31.1 - In Exercises 7–10, show that each function y =...

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