Explanation Given: The mass is 1-kg Stiffness of spring is 100 N-sec/m The damping constant is 0.2 N-sec/m Approach: 1. The linear second-order constant-coefficient differential equation a y ″ + b y ′ + c y = f ( t ) , here ( a ≠ 0 ) , with the case that f ( t ) = 0 , that is the equation a y ″ + b y ′ + c y = 0 is called the homogeneous equation. 2. To find the solution of second order homogeneous linear differential equation, some steps are given below. • Put y = e r t in the equation a y ″ + b y ′ + c y = 0 . • Get a equation of the form a r 2 + b r + c = 0 which is called the auxiliary equation or characteristic equation of the equation a y ″ + b y ′ + c y = 0 . • The characteristic equation is quadratic equation. Let r 1 and r 2 be two roots of the characteristic equation, let y 1 ( t ) and y 2 ( t ) be the pair of solutions. If the roots of the characteristic are real and distinct then general solution of the differential equation is given as, y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) . • If the characteristic equation has repeated roots then general solution of the differential equation is given as, y ( t ) = c 1 y 1 ( t ) + c 2 t y 2 ( t ) . • If the characteristic equation has complex roots say α ± i β then general solution of the differential equation is given as, y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) y ( t ) = c 1 e α t ( cos β t + i sin β t ) + c 2 e α t ( cos β t − i sin β t ) = e α t ( c 1 cos ( β t ) + c 2 sin ( β t ) ) The generalized equation mass-spring system with the effects of damping on free variables is given by m y ″ + b y ′ + k y = 0 … ( 1 ) Here, m is mass of the body, b is damping constant, and k is stiffness of the spring. Calculation: As the mass is pushed rightward from the equilibrium position with velocity of 1 m/sec , the initial conditions are y ( 0 ) = 0 y ′ ( 0 ) = 1 Here, y ( 0 ) is the initial position of the mass and y ′ ( 0 ) is the initial velocity of the mass. Substitute 1 for m , 100 for k , and 0.2 for b in equation ( 1 ) . y ″ + 0.2 y ′ + 100 y = 0 The above equation can be written as, y ″ + 2 10 y ′ + 100 y = 0 y ″ + 1 5 y ′ + 100 y = 0 5 y ″ + y ′ + 500 y = 0 … ( 2 ) Put e r t for y ( t ) in equation ( 2 ) . e r t ( 5 r 2 + r + 500 ) = 0 As e r t ≠ 0 , this implies that 5 r 2 + r + 500 = 0 The auxiliary equation of the above is 5 r 2 + r + 500 = 0 . Solve the auxiliary equation. r = − 1 ± 1 − 4 ( 5 ) ( 500 ) 2 ( 5 ) = − 1 ± i 1 − 10000 10 = − 1 10 ± i 9999 10 The roots are distinct and complex. r 1 = − 1 10 + i 9999 10 r 2 = − 1 10 − i 9999 10 . Therefore, the general solution of the equation ( 2 ) is, y ( t ) = e − t 10 ( A cos ( 9999 10 t ) + B sin ( 9999 10 t ) ) … ( 3 ) Here A , B are constants. Put 0 for y ( 0 ) in equation ( 3 ) . y ( 0 ) = e 0 ( A cos ( 0 ) + B sin ( 0 ) ) 0 = A ( 1 ) + B ( 0 ) A = 0 Calculate the first partial derivative of the equation ( 3 ) with respect to t

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Chapter 4.9, Problem 11E

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The time when the maximum displacement will be attained by the mass to the right.

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Chapter 4.9, Problem 10E

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The time when the maximum displacement will be attained by the mass to the right.

Solution Summary: The author explains that the maximum displacement will be attained by the mass to the right after 0.156mathrmsec

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To find: The time when the maximum displacement will be attained by the mass to the right.

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Chapter 4 Solutions

To find:

The time when the maximum displacement will be attained by the mass to the right.