Thursday

April 24, 2014

April 24, 2014

Posted by **annie** on Thursday, May 21, 2009 at 9:24pm.

A)Determine the frequency of the oscillation, assuming no damping, if k= 125N/m and M= 250g. (I got the answer to be 5.03 hz)

B)It is observed that after 51 oscillations, the amplitude of the oscillation has dropped to one-half of its initial value. Estimate the value of (lambda), using A(e^t*lambda)cos(w't)

C)How long does it take the amplitude to decrease to one-quarter of its initial value?

- physics spring -
**Damon**, Thursday, May 21, 2009 at 9:44pmequivalent spring constant = 2k = 250 N/m

w = sqrt(k/m) = srt(250/.25) = sqrt(1000)=10 sqrt 10

2 pi f = w = 10 sqrt 10

f = (10/2pi)sqrt 10 = 5.03 check

e^Lt = .5

T = 1/f = .1987 seconds

51 T = 10.13 s

so

e^L(10.13) = .5

10.13 L = ln .5 = -.693

L = -.0684

Now you have L

e^-.0684 t = .25

etc

- physics spring -
**annie**, Friday, May 22, 2009 at 12:05ambut how do you find A (i'm assuming amplitude)?

- physics spring -
**Damon**, Friday, May 22, 2009 at 5:47amIt did not ask for A. You can not find A. All you can find is the ratio of A to the original A.

A/Ao = e^Lt

- physics spring -
**annie**, Friday, May 22, 2009 at 3:29pmthe equation to use is more specifically x(t)=A(e^t*L)cos(w't)

does that change it? because i'm still not getting the right answer

- physics spring -
**sara**, Saturday, March 10, 2012 at 10:20amThe answer is right but it should be positive since the formula as I know it is e^(-L*t)

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