Explanation Formula used: The governing equation of motion is m d 2 y d t 2 + δ d y d t + k y = 0 . The Hooks Law states as: F = k x . Theorem used: “If r 1 and r 2 are the two real and unequal roots of the auxiliary equation a r 2 + b r + c = 0 , then y = c 1 e r 1 x + c 2 e r 2 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 weight ( w ) of object is 8-lb stretch a spring with displacement ( x ) 4 ft . 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 w and x in F = k x . 8 = 4 k k = 2 Calculate the mass, w = m g Where, g is acceleration due to gravity and its value in term of British system in earth is 32 ft/sec 2 . 8 = 32 m m = 1 4 s l u g The instantaneous velocity δ = 1.5 . And the external force is f ( t ) = 6 + e − t . The governing equation of motion is: m d 2 y d t 2 + δ d y d t + k y = f ( t ) With initial conditions y ( 0 ) = − 2 and y ′ ( 0 ) = 3 ft/s . Substitute the value of m , k , f ( t ) = 6 + e − t and δ in m d 2 y d t 2 + δ d y d t + k y = f ( t ) . 1 4 d 2 y d t 2 + 1.5 d y d t + 2 y = 6 + e − t d 2 y d t 2 + 6 d y d t + 8 y = 24 + 4 e − t The auxiliary equation for the above differential equation is: r 2 y + 6 r y + 8 y = 0 ( r 2 + 6 r + 8 ) y = 0 ( r 2 + 4 r + r + 8 ) = 0 ( r + 2 ) ( r + 4 ) = 0 . To calculate the value of r by split the middle term. The roots of the above equation are ( r + 2 ) ( r + 4 ) = 0 . The value of r = − 2 , − 4 . The general solution to the above differential equation is, y c ( t ) = c 1 e − 2 t + c 2 e − 4 t Solve the nonhomogeneous solution of the form y p = A + B e − t . Differentiate the equation y p = A + B e − t is as follows: y ′ p ( t ) = − B e − t Again, differentiate the above equation y ′ p ( t ) = − B e − t is as follows: y ′ ′ p ( t ) = B e − t Substitute the derivative of y p in the given equation d 2 y d t 2 + 6 d y d t + 8 y = 24 + 4 e − t and find that A and B must satisfy the equation. B e − t − 6 B e − t + 8 ( A + B e − t ) = 24 + 4 e − t 8 A + 3 B e − t = 24 + 4 e − t Compare both coefficients of the above equation is as follows: 8 A = 24 A = 3 To calculate the value of B. is as follows: 3 B e − 2 t = 4 e − 2 t 3 B = 4 B = 4 3 Now, substitute the value of A and B in y p = A + B e − t . y p = 3 + 4 3 e − t The general solution of the nonhomogeneous equation is: y ( t ) = y c ( t ) + y p ( t ) Substitute the value of y c ( t ) = c 1 e − 2 t + c 2 e − 4 t and y p = 3 + 4 3 e − t in y ( t ) = y c ( t ) + y p ( t )

14th Edition

ISBN: 9780134665672

Author: Hass

Publisher: PEARSON

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