Answer:
5.9*10^{-4}m
Explanation:
to find the uncertainty of the displacement it is necessary to compute the uncertainty for the angular frequency:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{0.40s}=15.707rad/s\\\\\frac{d \omega}{\omega}=\frac{dT}{T}\\\\d\omega=\omega \frac{dT}{T}=(15.707rad/s)\frac{0.020s}{0.40s}=0.785rad/s[/tex]
then, you can calculate the uncertainty in angular displacement:
[tex]\theta=\omega t\\\\\frac{d\theta}{\theta}=\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}\\\\d\theta=\theta\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}=0.0422[/tex]
finally, by using:
[tex]y=Acos(\omega t)\\\\dy=dAcos(\omega t)d(\omega t)=(dA)cos(\theta)d\theta=(0.002m)cos(0.785)(0.0422)\\\\dy=5.9*10^{-4}m[/tex]
The uncertainty of her displacement in mm is : 0.59 mm
First step : determine the uncertaintiy of the angular frequency
w = [tex]\frac{2\pi }{T}[/tex] = [tex]\frac{2\pi }{0.40} = 15.707 rad/s[/tex]
[tex]\frac{dw}{w} = \frac{dT}{T}[/tex]
therefore :
dw = 0.785 rad/s
Next step : determine the uncertainty of the angular displacement
θ = wt
dθ / θ = [tex]\sqrt{(\frac{dw}{w} )^2 + (\frac{dt}{t} )^2}[/tex]
therefore :
dθ = 0.0422
Final step : determine the uncertainty of displacement
y = Acos(wt)
dy = dAcos(wt)d(t) = (dA)cosθdθ
= ( 0.002m )cos (0.785)(0.0422)
= 5.9 * 10⁻⁴ m = 0.59 mm
Hence we can conclude that the uncertainty of her displacement in mm is : 0.59 mm
Learn more about displacement : https://brainly.com/question/321442
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