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A student gets a 2.9 kg mass to oscillate up and down on bottom of a light vertical spring by pulling her hand up and down on the top end of the spring. The spring is a real spring with a spring constant of 93 N/m and a damping constant of 0.7 N sec/m.

(b) The student now stops moving her hand and the mass slowly comes to rest. How long after she stop shaking her hand will it take for the amplitude of the mass to reach one half its maximum amplitude?

Tried the formula 2pi(m/k)^2 but not working, don't know what to do... help

  • Physics - ,

    Damping is exponential.

    F(h)= a * e^-bt/2m * sin(wt+ phase)

    with b the damping constant, which comes from a differential equation. I think you have the units wrong. The units should be mass/time

    Hmmm. I see you have N sec/m which is mass*m/sec^2*sec/mass...= sec/kg Ok, you can use those units.

    solution:

    1/2= e^-.7/5.8 t
    take the ln of each side.
    -.693=.7/5.8 t solve for t

  • Physics - ,

    You need the formula that relates the damping constant "b" (retarding force divided by speed) to the exponential damping RATE (lambda) for damped harmonic oscillations. This must have been in your reading material somewhere.

    If not, see "underdamped oscillations" in
    http://physics.ucsc.edu/~josh/6A/book/harmonic/node18.html

    The exponential damping time constant is
    lambda = b/2m = 0.12 sec-1

    e^-lambda*t = 1/2 when
    -lambda*t = -0.693
    t = 5.8 seconds

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