Explanation Formula used: The newton’s second law is as follows: mass × acceleration = restoring force + external force . The Hooks Law is as F = k x . 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 = 0.125 slugs So, the newton’s second law is as follows: mass × acceleration = restoring force + external force m × d 2 x d t 2 = − 4 x + 4 sin 2 t m d 2 x d t 2 + 4 x = 4 sin 2 t . Substitute the value of m in m d 2 x d t 2 + 4 x = 4 sin 2 t . 0.125 d 2 x d t 2 + 4 x = 4 sin 2 t d 2 x d t 2 + 32 x = 32 sin 2 t The auxiliary equation is: m 2 + 32 = 0 m 2 = − 32 m = ± 4 2 j So, the general solution to the above differential equation is x c = e 0 t ( c 1 sin 4 2 t + c 2 cos 4 2 t ) = c 1 sin 4 2 t + c 2 cos 4 2 t To 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 + 32 x = 32 sin 2 t and find that A and B must satisfy the equation. − 4 A sin 2 t − 4 B cos 2 t + 32 ( A sin 2 t + B cos 2 t ) = 32 sin 2 t ( − 4 A + 32 A ) sin 2 t + ( − 4 B + 32 B ) cos 2 t = 32 sin 2 t Equating the coefficients of similar terms: − 4 A + 32 A = 32 A = 8 7 − 4 B + 32 B = 0 B = 0 Now, substitute the value of A and B in x p = A sin 2 t + B cos 2 t

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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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