1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: SHM and hookes law

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data
    The diagram show a student before and after she makes a bungee
    jump from a high bridge above a river. One end of the bungee cord, which is of
    unstretched length 25 m, is fixed to the top of a railing on the bridge. The other end of
    the cord is attached to the waist of the student, whose mass is 58 kg. After she jumps,
    the bungee cord goes into tension at point P. She comes to rest momentarily at point R
    and then oscillates about point Q, which is a distance d below P.

    2. Relevant equations
    The bungee cord behaves like a spring of spring constant 54Nm–1.
    Calculate the distance d, from P to Q, assuming the cord obeys Hooke’s law.

    3. The attempt at a solution
    What I can’t see is why this works for this distance when the cord stretches from P all the way down to R. Also only at P is the force just due to the weight (mg). I am assuming it has something to do with Q being the equilibrium position of the oscillation. Any pointers for my thought processes gratefully received.
    Last edited: Dec 16, 2012
  2. jcsd
  3. Dec 16, 2012 #2
    I think I get it now. Q is the position that the girl on the rope would hang if stationary once all the energy has dissipated. Hence at Q mg=T and the extension of the bungee cord if a force of the girls weight were added would be from upstretched length to this new length
  4. Dec 16, 2012 #3
    There is a further question that although I can apply the formulas I still struggle to see why this works.
    Use your answers from parts (a) and (b)(i) to show that the amplitude A, which is the
    distance from Q to R, is about 25 m.
    Why can we use these values? Is it simply that we can say that on return to the point P after the first oscillation she will arrive there with the same velocity as she had when she was originally fell to P through the 25m distance. The speed at P being calculated from KE=GPE (v=sqrt(2gh). Then from conservation of energy if we ignore any dissipation of the energy the girl will bounce back up and be at the same speed at P again when she returns to this point moving vertically upwards?
  5. Dec 16, 2012 #4
    OK the simple harmonic oscillation of a mass on a spring is a little more tricky than some, but still accessible to high school physics.

    Remember the definition of SHM?

    The motion such that there is always a restoring force and consequent acceleration directed towards a fixed point and proportional to the distance from it.

    So consider your jumper.
    Starting from the end of the bungee slack, (your point P) she is moving under the effect of two accelerations. Gravity and the bungee.

    Initially gravity is greater than that provided by the bungee, but as the bungee stretches its restoring force increases until the point Q where it is exactly equal to that of gravity.

    mg = ke where e = PQ

    This is your first equation.

    Note that from P to Q the acceleration and restoring force is directed downwards towards Q.

    At Q the jumper has achieved some velocity vq and thereofre keeps going, still under the influence of gravity and the bungee tension.
    However the tension continues to increase as the bungee stretches further and the tension is now greater than gravity and opposing it. Thus the resultant is still directed towards Q, as required by our SHM definition.

    Measuring distance vertically from Q as y with downwards positive we find for some point B, y below Q.

    extension of bungee = y+e and the tension = k(y+e)

    Thus resultant = mg-k(y+e) = mg - ky - ke = -ky since ke=mg

    Hence mg = -ky


    acceleration = -k/my = -ω2y, where ω2 = k/m

    So we have proved SHM about Q and can use the SHM formule

    Period = 2∏/ω = 2∏√(m/k)

    Also since mg=ke

    Period = 2∏√(e/g)

    Does this help?
  6. Dec 16, 2012 #5
    Thank you so much, this makes a great amount of sense to me.
  7. Dec 16, 2012 #6
    Glad it helped.

    Would you like to consider an energy balance for the system?

    This is slightly more complicated since there are several different energies involved. Can you spot them?
  8. Dec 16, 2012 #7
    I would love to consider the energy balance of the system.
    gravitational potential energy determined by the distance between P and R, however the GPE in the fall through the distance of 25m to get to P was used to calculate the speed at P.

    elastic energy this is zero at P and increases untill maximum at R then decrease as she returns back up to P.

    and kinetic energy which must decreas from P to R as the speed decreases and the KE turns to the elastic potential energy. is it also true to say while descending from P to R the GPE is decreasing and turning into KE and then as she moves upwards from R back to P the KE and elastic turn into GPE all GPE at the maximum height.
    This question came from an AQA past paper

    I really appreciate the help and I suppose I should also think about the direction and magnitude of the acceleration.
    Aint physics brilliant!
  9. Dec 16, 2012 #8
    The total gravitational potential energy available to the system is

    The mass times the total fall from the railing to R (times g).

    This is distributed into the vibrating system as

    (1)strain energy in the bungee, (potential energy)
    (2)kinetic energy of the person
    (3)gravitational (potential) energy of the person.

    If you take R as a zero point for energy purposes you can see that at R (2) and (3) are zero and at P (2) is zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook