ETScustomer 20170928 21:04:43  Here\'s to see why the frequency is not always changing and rule out (E).\r\nWe have the differential equation:\r\n\r\nThat is,\r\n\r\nThe general solution is\r\n\r\nAssuming we\'re not too damped, the new angular frequency is less than the old . Thus the period is a constant in time that is larger than .
ETScustomer 20170928 21:06:16 
Woah, that looked nice in the preview, but good luck reading that here.\r\n

 
GRE1995 20160926 11:09:10  A physical way to view this problem would be to consider a pendulum displaced from equilibrium by a set distance, and then returning to it. The presence of damping should make the pendulum slower at every point, and hence the answer is (B) or (E). (Notice that for a simple pendulum period independent of damping, so it does not make a difference that the damped pendulum would get to a lower height). Not sure how to rule out E, but you could do it if you recalled the graph  
someone 20111105 10:30:19  If you forgot it, you can quickly derive it
. You know solutions will be of the form substitute and write down the quadratic
Solutions to this are
The first term is the exponentially decaying envelop and the second term would be imaginary which causes the oscillations (check that for b=0 it goes to ). So we know that $b^2 4km <0 \Rightarrow $ frequency is smaller therefore, time period is larger.
 
Steve 20110824 14:36:27  I missed this problem the first time (put B) because I think I was thinking about instead of .
Can anyone confirm that a damping force (where x=0 is the equilibrium position of the oscillation) would decrease the period?
physick 20111009 07:51:22 
The force you describe is not a damping force, it is a restorative spring force. The form is the same as Hooke's Law, .

 
Kabuto Yakushi 20100905 09:39:38  One can see at once that the answer must be either (B) or (A). We know that when a linearly resistive force is introduced to harmonic oscillations the period: will not be same as the natural period. Since (B) and (A) cannot both be false it must be one or the other, this takes out (C), (D), and (E). Common sense tells us that the oscillations won't speed up when a resistive force is applied! Therefore the answer is (A). QED. Not very rigorous, but saves time on the GRE.  
joshfr3 20100527 17:38:16  For the record, the original answer should be reworded to exclude the part that states "As the amplitude shrinks, the period increases." This would imply that the frequency is constantly changing, which, it is not. The way the answer is worded makes testers believe that (E) could be a potential answer: (E) constantly changing period.
maryami 20110331 04:29:37 
This is a false comment,period of oscillator is not related to amplitude of oscillations,as T=2pi sqrt(m/k)

pam d 20110911 17:32:06 
Actually joshfr3 is 100% correct here. Maryami misread his comment. Furthermore, the equation maryami supplied only applies to nondamped systems.

 
Kentai 20091105 22:08:52  with the equation change from x'' = w^2 to x'' = bx'  w^2 x, w must become smaller to compensate the bx' term, hence the period increase.  
jmason86 20091004 13:26:22  No one has addressed the other options that ETS gives us. I thought that (D) was correct because of limits. If b> that implies an infinite force acting against the oscillator, which I would imagine would completely destroy the period because it would just stop moving. If b>0 then you have no damping effect at all and it should behave like an ideal SHO with a period of .
Why is this not correct?!
kroner 20091009 19:27:03 
The period does increase with increasing b, but it's not a linear relationship. See the formula provided by sharpstones below.

 
phun_lover 20090929 17:56:23  This question is somewhat confusing since a damped oscillator doesn't have a welldefined period, in the sense that it's position isn't a periodic function of time. However, it's position can be written as a decaying exponential times a periodic function (if the damping is in the correct range), and so what they want here is the period of that periodic factoreasily solved for from the corresponding differential equation.  
cobrachi 20081031 22:14:48  It's simple to think of this conceptually. The new force is in the opposite direction of the velocity so it will act against the motion of the particle. Thus, it decreases the angular frequency and since T=2pi/w > a smaller w results in a larger T.  
greenfruit 20081031 08:36:12  Question about the soln: Why can we conclude that as amplitude shrinks period increases?  
gn0m0n 20081020 01:22:45  To echo a couple of questions and hopefully clarify them: do we mean that the period is greater than when it was undamped (ie, but still constant once it is damped) or do we mean it is changing in time once the damping is applied?
gn0m0n 20081020 01:24:19 
I'd like to ask the same thing about the amplitude.

 
sharpstones 20070402 19:45:51  Just to tex it out. The general solution to a damped oscillator is: where where is the damping term and is the frequency of the solution
if omega is real (which is the underdamped case: ) you will have oscillations which do in fact have a constant period but will have decreasing amplitude from the . clearly the frequency will be less then the original frequency so the period will be greater.
sharpstones 20070402 19:46:55 
that should be

blah22 20080214 11:02:07 
I'm confused. Why do you say you will have oscillations which have a constant period and in the next sentence say the frequency will clearly be less?

gt2009 20090628 13:20:30 
Both periods are constant, but the damped period is greater than the undamped period.

 
huanggyellow 20070320 07:28:47  How about choice (E)? Surely the period is constantly changing (increasing)?
mhas035 20070321 21:42:23 
The frequency of an underdamped oscillator is omega = omega{undamped}*sqrt(1b.^2/4mk), i.e. smaller than the undamped frequency, and constant.

 