{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Key_to_ME130_F08_Exam1

Key_to_ME130_F08_Exam1 - Dr Saxena ME130_F08 ME130 APPLIED...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Dr. Saxena ME130_F’08 1 ME130 – APPLIED ENGINEERING ANALYSIS Key to Exam #1: Fall 2008 – Section 02: #1 A 5-kg mass is hanging from a spring with a spring constant K = 30 N/m, and there is a dash pot attached to the mass with a damping coefficient C = 25 Kg/sec. The initial conditions are: y(0) = 1 m, and y (0) = 0. (a) Derive the governing differential equation for the displacement y(t) of this mass (b) Solve the equation to determine the displacement y(t). (c) Calculate the percent reduction in the amplitude in 1 sec. Solution: (a) From the free body diagram F = F I R S R D = 0, where R D = Spring force = K y(t), K = spring constant R D = Viscous (damping) force = C dy/dt, C = damping coefficient F I = Inertia force = m d 2 y/dt 2 So, the governing differential equation becomes 2 2 d y dy M + C + K y = 0 dt dt (Note: the weight is offset by initial stretch of the spring) (b) This is a second order homogeneous differential equation, with constant coefficients, with the initial conditions given as: y(0) = 1 m, and y’(0) = 0 The corresponding characteristic equation is: m 2 + am + b = 0, where a = C/M = 25/5 = 5, and b = K/M = 30/5 = 6. The two roots of this equation are : ( ) ( ) ( ) ( ) 3 6 4 5 2 1 2 5 b 4 a 2 1 2 5 m 2 6 4 5 2 1 2 5 b 4 a 2 1 2 a m 2 2 2 2 2 1 = × = = = × + = + = Therefore, the general solution of this equation is expressed as t 3 t 2 t m t m e B e A e B e A ) t ( y 2 1 + = + = Using the given initial conditions, the two constants are A = 3, and B = 2. Therefore, the specific solution of the differential equation is t 3 t 2 e 2 e 3 ) t ( y = (c)
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern