physics spring
posted by annie on .
A glider on an air track is connected by springs to either end of the track. Both springs have the same spring constant,k , and the glider has mass M.
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 onehalf 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 onequarter of its initial value?

equivalent 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 
but how do you find A (i'm assuming amplitude)?

It 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 
the 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 
The answer is right but it should be positive since the formula as I know it is e^(L*t)