Question

A student is studying simple harmonic motion of a spring. She conducts an experiment where she measures the amplitude and period of an undamped system to be 23 ± 2 mm and 0.40 ± 0.020 seconds, respectively. Using the equation for displacement as a function of time y(t) = Acos(ωt), what is the uncertainty of her displacement calculation in mm for t = 0.050 ± 0.0010 seconds?

Answer 1

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:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.40s}=15.707rad/s\\\\\frac{d \omega}{\omega}=\frac{dT}{T}\\\\d\omega=\omega \frac{dT}{T}=(15.707rad/s)\frac{0.020s}{0.40s}=0.785rad/s$

then, you can calculate the uncertainty in angular displacement:

$\theta=\omega t\\\\\frac{d\theta}{\theta}=\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}\\\\d\theta=\theta\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}=0.0422$

finally, by using:

$y=Acos(\omega t)\\\\dy=dAcos(\omega t)d(\omega t)=(dA)cos(\theta)d\theta=(0.002m)cos(0.785)(0.0422)\\\\dy=5.9*10^{-4}m$