Explanation Formula used: The governing equation of motion is m d 2 y d t 2 + δ d y d t + k y = f ( t ) . The Hooks Law is as F = k x . 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 . The general solution y = y ( x ) to the nonhomogeneous differential equation a y ″ + b y ′ + c y = G ( x ) has the form y = y c + y p , where the complementary solution y c is the general solution to the associated homogeneous equation a y ″ + b y ′ + c y = 0 , and y p is any particular solution to the nonhomogeneous equation a y ″ + b y ′ + c y = G ( x ) . Calculation: The eight of the spring is 4 .00 lb with initial stretch ( x ) is 0.125 ft, the spring is pulled by 3.0 in longer than the equilibrium length and then released, and the external force acting on the spring is 4 sin 2 t . From the Hooks Law, F = k x . Where k is spring constant, x is displacement of spring and F is force applied on the spring. Substitute the value of F and x in F = k x . F = k x 4.00 = k × 0.125 k = 4 0.125 k = 32 lb/ft Now, to get the mass of object: F = m a 4.00 = m × 32 m = 4 32 m = 0.125 slugs So, the governing equation of motion is as follows: m d 2 y d t 2 + δ d y d t + k y = f ( t ) . Substitute the value of m , δ and k in m d 2 y d t 2 + δ d y d t + k y = f ( t ) . 0.125 d 2 x d t 2 + 1 × d x d t + 4 x = 4 sin 2 t d 2 x d t 2 + 8 d x d t + 32 x = 4 sin 2 t The auxiliary equation is: m 2 + 8 m + 32 = 0 Compare the above equation with a x 2 + b x + c = 0 . Here a = 1 , b = 8 , c = 32 . Substitute the value of a = 1 , b = 8 , c = 32 in the quadratic formula x = − b ± b 2 − 4 a c 2 a . m = − 8 ± 8 2 − 4 × 1 × 32 2 × 1 = − 8 ± 64 − 128 2 = − 8 ± − 64 2 = − 4 ± 4 j So, the general solution to the above differential equation is x c = e − 4 t ( c 1 sin 4 t + c 2 cos 4 t ) Solve the nonhomogeneous solution of the form, x p = A sin 2 t + B cos 2 t . Substitute the derivative of x p in the given equation d 2 x d t 2 + 8 d x d t + 32 x = 4 sin 2 t and find that A and B must satisfy the equation. − 4 A sin 2 t − 4 B cos 2 t + 8 ( 2 A cos 2 t − 2 B sin 2 t ) + 32 ( A sin 2 t + B cos 2 t ) = 4 sin 2 t ( − 4 A + 32 A − 16 B ) sin 2 t + ( − 4 B + 16 A + 32 B ) cos 2 t = 4 sin 2 t ( 28 A − 16 B ) sin 2 t + ( 16 A + 28 B ) cos 2 t = 4 sin 2 t Equating the coefficients of similar terms: 28 A − 16 B = 4 16 A + 28 B = 0 Simplify the above obtained equations as shown below. 7 A − 4 B = 1 … ( 1 ) 4 A + 7 B = 0 … ( 2 ) Consider 4 A + 7 B = 0 . 4 A + 7 B = 0 4 A = − 7 B A = − 7 4 B Substitute the value of A = − 7 4 B in 7 A − 4 B = 1

11th Edition

ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Question

Chapter 31.10, Problem 16E

To determine

The equation of resulting motion if a damping force numerically equal to the velocity is present.

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Chapter 31.10, Problem 15E

Chapter 31.10, Problem 17E

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Chapter 31 Solutions

Basic Technical Mathematics

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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