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Homework Help: Vibrating spring

  1. Jun 19, 2006 #1
    Hi i need some help with a differential equations problem. The question is:-
    (w/g)y(double prime) + ky(single prime)+cy=F(t)
    w= the weight of the object attached to the spring
    k= damping factor
    c=spring constant
    F(t)= the external force on the system.
    A 64 pound weight is attached to the end of the spring. After reaching the equilibrium position the spring is stretched one foot beyond the equilibrium position. The weight is then released and as it is released it is struck a downward blow giving it an initial velocity of 2 ft/sec. Take the moment the weight is released and struck as time zero. At time zero a periodic external force given by F(t) = (1/2)cos(4t) pounds begins acting on the system. t is time in seconds. Consider the damping factor to be negligible, i.e., take k to be zero. The spring constant is 32. Find the function giving y, the position of the bottom of the weight as a function of time given in seconds.
    if some one could help me with this problem i would really appreciate it.
  2. jcsd
  3. Jun 21, 2006 #2
    first of all im somewhat confused in terms of the notation
    typically c is the damping constant and k is the spring constant
    I will use this notation in my discussion here...

    the EOM can be written as

    y'' + (k/m)*y = F/m*cos(wd*t)

    here k is the spring constant, m is the mass (w/g), and wd is the driving frequency.

    Notice that the system has two complex roots with entirely imaginary parts. The solution to the homogeneous equation...

    y'' + (k/m)*y = 0 takes the form

    y = c1*cos(w*t) + c2*sin(w*t)

    where c1 and c2 are arbitrary constants that are determined by the initial conditions and w is the natural frequency of system which is given by (k/m)^1/2. However, we cannot apply the boundary conditions yet to solve for c1 and c2. We must first consider the driving term.

    Notice that for this problem w = sqrt(k/m) = sqrt(k*g/w) = sqrt(16) = 4
    and wd = 4. This is not good, this means the system is going to be highly unstable because the driving frequency is the same as the natural frequency. Thus the amplification factor is infinit. So by using the method of undetermined coefficients we need to look for a solution of the form
    A*cos(wd*t) + B*sin(wd*t) however this is not entirely valid because the new particular solution is not independent of the homogeneous solution. Thus we are really looking for a new solution of the form y = t*(A*cos(w*t)+B*sin(w*t)).

    Now we have to solve

    y'' + 16*y = 1/4*cos(4*t)

    use y = t*(A*cos(w*t)+B*sin(w*t)).
    and differentiate twice to find and expression for y''

    Now solve for A and B such that the coefficient of the sin term is zero
    and the coefficient of the cosine term is 1/4

    You will have 2 eqns and 2 unknowns such that

    F(A,B)*cos(4*t) + G(A,B)*sin(4*t) = 1/4*cos(4*t)

    So solve F(A,B) = 1/4 and G(A,B) = 0
    this will give you the particular solution.

    The entire solution has the form y = c1*cos(4*t) + c2*sin(4*t) + Yp

    where Yp = t*(A*cos(4*t)+B*sin(4*t)) where you have solved for A and B above

    Now with A and B known plug into

    y = c1*cos(4*t) + c2*sin(4*t) + Yp

    and now apply the initial (boundary conditions) which are the initial displacement and initial velocity.

    notice that c1*cos(4*t) + c2*sin(4*t) can be written as R*sin(4*t+p)

    where R = sqrt(c1^2+c2^2) and p = atan(c1/c2)

    In the end you should have a sin wave for 1 term and then another sne wave with a secular term (term that grows with time)

    I would expect a solution of the form

    (C+D*t)*sin(4*t + E)

    where C D and E are numbers.

    This problem is significantly ugly because the driving and natural frequencies are the same. Essentially this is like a person pushing you on a swing at the same frequency at which the swing would naturally osciallate. If we neglectthe air resistance and friction in the links this basically states that after enough pushes you would end up going in circles around the pole holding the swing!
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