# Homework Help: Real Spring

1. Nov 16, 2009

### bricker9236

1. The problem statement, all variables and given/known data

A student gets a 3 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 89 N/m and a damping constant of 0.5 N sec/m.

2. Relevant equations

(a) At what approximate frequency should the student move her hand up and down to get the maximum motion from the mass with the minimum motion of her hand?
HELP: What approximately driving force frequency gives the greatest motion for a driven harmonic oscillator?
HELP: Notice that we're asking for the frequency, f, not the angular frequency, ω.

I got this answer correct. I took the Fo=1/T= (1/2pi)(sqrt(k/m))=(1/2pi)(sqrt(89/3))=.866
Now part B is what i am having trouble with.

(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?
HELP: What is the formula for the maximum applitude of a object on a damped spring as a function of the damping constand and the mass of the object?
HELP: Remember from math class that if y = ex, then ln(y) = x.
HELP: If you would like to just type in the formula for this answer instead of calculating it out, you could look at some of the functions available in the "Instructions" link below

Now i noticed that the help gave a formula that I have yet to see in the book which really threw me off. y=ex then ln(y)=x. That is not in the section of damped oscillations or antyhing around that. if anyone knows what that equation is trying to infer let me know. because i tried looking for the max amplitude formula and i got

A=Aoe superscript -bt/2m

3. The attempt at a solution

Stated above.

2. Nov 16, 2009

### Delphi51

Hi Bricker. I can help you with "y=e^x then ln(y)=x". That is just a mathematical statement that the logarithm is the inverse of the exponential function. As an example in base 10, since 100 = 10², it tells you that log(100) = 2. It is a handy way to figure out the exponent in an exponential formula.

I don't quite know how to solve (b). There seems to be a complicated relationship between the damping constant you are given and the corresponding constant in the exponential decay formula. Check out this link:
http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html#c1
I think your damping constant is the "c" in the link. You have to figure out the "q" so you can deduce when e^(qt) is 0.5. Hope that equation relating c and q looks familiar to you! Some other references are using a weird greek letter instead.

3. Nov 21, 2009

### vishal007win

@bricker
try the energy approach dats makes life simpler...
energy pumped by the external force = energy dissipated by damper ---1st part of ur question

and
energy diff. from amplitude A to A/2 = energy dissipated by damper in time t

rate at which damper dissipate power =d*v*(dv/dt)