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Vertical SHM. Find f with A and drop height

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data

    A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height of 3.0cm above the top of the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 10cm. What is the oscillation frequency?

    2. Relevant equations

    f = 1/(2*pi)*sqrt(k/m) (frequency equation)

    Ug = m*g*y (gravitational potential)

    E = 1/2*k*A^2 (system energy equation)

    3. The attempt at a solution

    Known
    • A = 10cm = 0.1m
    • mass = m
    • spring constant = k
    • mass brings energy into the system
    • mass had 3cm (0.03m) displacement before contacting spring

    1.) use energy

    because I know A, I will use the system energy equation that includes it.

    0.005k = 1/2*k*(0.1^2)

    It seems that I will be able to solve for both k and m, but I'm going to have to use the information given about the block's 3cm displacement fall onto the spring.

    Maybe i could find out where the system's equilibrium position is relative to before the mass was added to it, but I don't think it could help me solve for k and m. I know it fell 3cm before contacting the spring, and that tells me information about its energy.

    I assume that all of the energy in the system = potential energy m started out with.

    .005k = m*9.81*0.03

    m = 50k/2943

    2.) solve for frequency

    1/(2*pi)*sqrt(k/(50k/2943) = 1/(2*pi)*sqrt(2943/50) = 1.22Hz

    The book answer is 1.83Hz


    In an attempt to get the right answer, I decided to follow a hunch that maybe I am supposed to add a distance to my Ug equation. With verticle SHM, the equilibrium position is placed a distance below the spring's natural equilibrium. I figured that if i could find this, maybe I could make up for the missing Hz.

    defining the position below the spring's natural equilibrium as ΔL,

    ΔL = (m*g)/k

    now my Ug equation looks like:

    Ug = m*g*(0.03 + (m*g)/k)

    This ends up becoming a quadratic:

    Ug = ((m*g)^2)/k + 0.03(m*g)

    if i define g as 9.81 i can clean it up a little to look like:

    Ug = (96.23m/k + 0.2943)*m


    The next step is to put it against the system's energy (1/2k*A^2), which was a pretty ugly process.

    With the two solutions for m: -0.0088k and 0.0058k, I tried the positive solution.

    My new answer is 2.08Hz, which is too large!

    I'm out of ideas now, is there something I'm doing wrong? Am I approaching the problem wrong?
     
  2. jcsd
  3. Jun 5, 2012 #2
    taking another look at the system...


    I found through FBD's that the energy brought in from the block could possibly be:

    (m*9.81*(0.03+ΔL)) - (1/2k*ΔL^2)

    That is, the Ug energy with the extra ΔL length added, subtracted by the Usp energy during the ΔL displacement.

    I figure that since the system adopts a new equilibrium point ΔL below it's natural one, the restoring force ought to be added in as well during that ΔL displacement. But whether this solves anything or proves that the ΔL was a pointless venture isn't clear to me yet.

    What the heck, I'll put that against the 1/2*k*(A^2) equation once again.

    With my CAS, it comes with m = 0.0075k as the positive solution.

    When I put that into the frequency equation I get 1.827Hz.

    Sorry guys, i think I solved this one :redface:
     
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