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Math Help - Diff Eq: Mechanical and Electrical Vibrations

  1. #1
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    Diff Eq: Mechanical and Electrical Vibrations

    Hi all, this is my first time posting here. I'm in diff eq at uni, and so far things have been going smoothly, except for this section! I don't have a strong physics background, so maybe that's what's tripping me up? In any case, I could really use some help with these problems:




    Problem: A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring, and then set in motion with a downward velocity of 2ft/sec, and if there is no damping, find the position u of the mass at any time t. Determine the frequency, period, ,amplitude, and phase of the motion.

    Answer:
    u = (1/4 (root 2))sin(8(root 2t)) - (1/12)cos(8(root 2t) ft, t in seconds;
    omega = 8(root 2) radians/seconds
    T = (pi/(4 root 2)) seconds
    R = (root (11/288)), or approx .1954 ft
    delta = pi - arctan (3/(root 2)), or approx. 2.0113



    Problem: Assume that the system described by the equation mu" + &u' + ku = 0 is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions.
    (The hint they give: determine all possible values of t for which u = 0)



    Problem: Assume that the system described by the equation mu" + (gamma)u' + ku = 0 is critically damped and that the initial conditions are u(0) = u(sub)0, u'(0) = v(sub)0. If v(sub)0 = 0, show that u approaches 0 as t approaches infinity, but that u is never zero. If u(sub)0 is positive, determine a condition on v(sub)0 that will lensure that the mass passes through its equilibrium position after it is released.

    answer:
    v(sub)0 < -(gamma)u(sub)0/2m




    Problem: The position of a certain spring-mass system satisfies the intiial value problem:

    (3/2)u" + ku = 0, u(0)=2, u'(0)=v

    If the period and amplitude of the resulting motion are observed to be pi and 3, respectively, determine the values of k and v.

    answer:
    k = 6, v = + or - 2(root 5)



    Problem: A cubic block of side l and mass density p per unit volume is floating in a fluid of mass density p(sub)0 per unit volume, where p(sub)0 > p. If the block is slightly depressed and then released, it oscillate in the vertical direction. Assuming that the viscous damping of the fluid and air can be neglected, derive the differential equation of motion and determine the period of the motion.
    (The hint they give: Use Archimedes' principle: An object that is completely or partially submerged in a fluid is acted on by an upward (buoyant) force equal to the weight of the displaced fluid).

    answer:
    (p)(l)(u") + (p(sub)0)(g)(u) = 0





    Any help given would be appreciated. I don't really understand how this relates, and I've yet to take a course in calc based physics (or any physics, for that matter). And I don't just need the answers (they're all in the back of the book), but the process to get to there.

    Thank you!
    --Rachel
    (oh, and I'm not sure how to make any Greek symbols or subscript or anything, sorry about that, if it's confusing :-/ )
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  2. #2
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    Re: Diff Eq: Mechanical and Electrical Vibrations

    First, "LesleyIlic" post has nothing to do with this topic- it is spam. Moderators, please delete it and ban LesleyIlic.

    Quote Originally Posted by tibetan-knight View Post
    Hi all, this is my first time posting here. I'm in diff eq at uni, and so far things have been going smoothly, except for this section! I don't have a strong physics background, so maybe that's what's tripping me up? In any case, I could really use some help with these problems:




    Problem: A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring, and then set in motion with a downward velocity of 2ft/sec, and if there is no damping, find the position u of the mass at any time t. Determine the frequency, period, ,amplitude, and phase of the motion.

    Answer:
    u = (1/4 (root 2))sin(8(root 2t)) - (1/12)cos(8(root 2t) ft, t in seconds;
    omega = 8(root 2) radians/seconds
    T = (pi/(4 root 2)) seconds
    R = (root (11/288)), or approx .1954 ft
    delta = pi - arctan (3/(root 2)), or approx. 2.0113
    Is this the solution that was given in the book? There are two physics principles involved here: f= ma (force is equal to mass times acceleration) and Hooke's law the force on a spring is proportional to the stretch or compression. We are told "A mass weighing 3 lb stretches a spring 3 in" so the proportion is 3lb/3 in= 1 lb/in or 3lb/(1/4 ft)= 12 lb./foot.

    A 3 lb weight has mass 3/32 (from f= ma again- weight is force and a= 32 ft/sec^2) so the differential equation involved is (3/32)d^2x/dt^2= -12x (the "-" is because if we stretch the spring, making x larger, the force is back against the stretch). The problem, as you wrote t, does not give an initial x value but putting t into the u gives x= -1/12. So apparently u(0)= -1/12 (one inch and x is in feet) and u'(0)= 2 ft/sec. Do you know how to solve u''= -12u with initial values u(0)= -1/12 and u'(0)= 2?



    Problem: Assume that the system described by the equation mu" + &u' + ku = 0 is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions.
    (The hint they give: determine all possible values of t for which u = 0)
    Can you find the general solution to the differential equation? Recall that for a second order linear equation with constant coefficients, if the solutions to the characteristic equation are a\pm bi then the general solution is of the for e^{ax}(Ccos(bx)+ Dsin(bx)) and will be "critically damped" if the characteristic equation has a "double" root- that is b= 0 and will be "overdamped" if the characterstic equation has two real roots.


    Problem: Assume that the system described by the equation mu" + (gamma)u' + ku = 0 is critically damped and that the initial conditions are u(0) = u(sub)0, u'(0) = v(sub)0. If v(sub)0 = 0, show that u approaches 0 as t approaches infinity, but that u is never zero. If u(sub)0 is positive, determine a condition on v(sub)0 that will lensure that the mass passes through its equilibrium position after it is released.

    answer:
    v(sub)0 < -(gamma)u(sub)0/2m
    Okay, if the equation is critically damped, the characteristic equation has a double root. What is the general solution to the differential equation? There is no "physics" involved here.




    Problem: The position of a certain spring-mass system satisfies the intiial value problem:

    (3/2)u" + ku = 0, u(0)=2, u'(0)=v

    If the period and amplitude of the resulting motion are observed to be pi and 3, respectively, determine the values of k and v.

    answer:
    k = 6, v = + or - 2(root 5)
    The characteristic equation is r^2+ k= 0 which has roots i\sqrt{k} so the general solution is u= Acos(\sqrt{k}t)+ Bsin(\sqrt{k}t)
    Now u(0)= 2 and u'(0)= v let you solve for A and B (in terms of v). What are the period and amplitude of u, in terms of k and v? Set the equal to [tex]\pi[/itex] and 3, respectively, and solve for k and v. There is no "physics" involved here.



    Problem: A cubic block of side l and mass density p per unit volume is floating in a fluid of mass density p(sub)0 per unit volume, where p(sub)0 > p. If the block is slightly depressed and then released, it oscillate in the vertical direction. Assuming that the viscous damping of the fluid and air can be neglected, derive the differential equation of motion and determine the period of the motion.
    (The hint they give: Use Archimedes' principle: An object that is completely or partially submerged in a fluid is acted on by an upward (buoyant) force equal to the weight of the displaced fluid).

    answer:
    (p)(l)(u") + (p(sub)0)(g)(u) = 0
    Let u be the depth to which the block sinks in the water. Then the water displaced (assuming u is not greater than l) is ul^2 and the much water has weight \rho_0 ul^2 and, by Archmede's principle, that is the upward force on the block: "mass times acceleration equals force" becomes \rho l^3 u''=  -g\rho_0 ul^2. Can you solve that differential equation?




    Any help given would be appreciated. I don't really understand how this relates, and I've yet to take a course in calc based physics (or any physics, for that matter). And I don't just need the answers (they're all in the back of the book), but the process to get to there.

    Thank you!
    --Rachel
    (oh, and I'm not sure how to make any Greek symbols or subscript or anything, sorry about that, if it's confusing :-/ )
    Follow Math Help Forum on Facebook and Google+

  3. #3
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    Re: Diff Eq: Mechanical and Electrical Vibrations

    A better reading??? depending on the type of gauge it shouldn’t really matter. as a personal preference id go with electronic; main reason being that you can tap into your vac. source at a closer distance making IMO a better reading since it’s an electrical signal to get from the sender to the gauge. If you where to go with a mech. gauge you have to run that vac. line all the way to where ever you place your gauge...thus making your reading a bit off from all the excess vac. line.
    Miniature Slip Rings
    Capsule Slip Rings Rotary joint
    to me it kinda fall into the same lines as a MBC, closer to the wastegate the better.
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