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November 29, 2014

November 29, 2014

Posted by **Jacob** on Tuesday, April 13, 2010 at 4:21pm.

(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 -
**bobpursley**, Tuesday, April 13, 2010 at 4:37pmDamping 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 -
**drwls**, Tuesday, April 13, 2010 at 4:46pmYou 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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