# physics spring

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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 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 -

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

• physics spring -

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

• physics spring -

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

• physics spring -

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

• physics spring -

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

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