Answer :
Answer:
b/2m = 0.001 s^-1
Explanation:
The formula for amplitude x = A e^-bt/2m
But we had the initial amplitude xi and the final amplitude reduced by friction as xf
Thus we have
final amplitude xf = Ae^-b(1000)/2m
and
Initial amplitude xi = A e^-bt/2m
Hence
5.508 = e^-b(1000)/2m --------------eqn 1
15.08 = e^(0)b/2m ---------------------eqn 2
equation 2 becomes
15.08 = e^(0) because any number multiplied by 0 = 0
and any number raise to the power of 0 = 1
Hence dividing xf/xi
We get
5.508/15.08 = e-b(1000)/2m
ln(5.508/15.08) = -1000b / 2m
-1.003 = -1000b / 2m
b/2m = 0.001 s-1
Given a pendulum with a certain height and angle, we have that the value of [tex]b/2m[/tex] will be [tex]0.001 s^{-1}[/tex]
Knowing that in the pendulum equation we find:
[tex]x_i = A e^{-bt/2m}[/tex]
[tex]x_f = Ae^{-b(1000)/2m}[/tex]
Where:
- [tex]x_i[/tex] corresponds to the initial amplitude
- [tex]x_f[/tex] corresponds to the final amplitude
We know that the values of X are equal to the values of the angles, so we can rewrite the two equations as:
[tex]5.508 = e^{-b(1000)/2m}\\15.08 = e^{(0)b/2m}[/tex]
With these values and solving the equation we find that:
[tex]5.508/15.08 = e^{-b(1000)/2m} \\ln(5.508/15.08) = -1000b / 2m\\-1.003 = -1000b / 2m\\b/2m = 0.001 s^{-1}[/tex]
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