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? Round your answer to 2 decimal places for entry into eCampus. Do not enter units. Example: 1.23

Answer 1

Step-by-step explanation:

y(t) = Acos(ωt)

taking log on both the sides

lny = lnA + lncos(ωt)

differentiating on both sides

`(dy)/(y) = (dA)/(A) +sin\omega t(d\omega t)/(cos\omega t)`

% change in y = % change in A + sinωt.% change in ωt

% change in A =
`(2)/(23) * 100` = 8.7 %

% change in ωt = % change in ω + percentage change in t

= % change in ω =
`(1)/(.40)` %

= 2.5 %

percentage change in t =
`( .001 )/( .05 )* 100`

= 2 %

% change in y = 8.7 % + 2.5 % + 2 %

= 13.2 %

y = Acos(ωt

= 23 cos
`(2\pi)/(.4)*.05`

= 23 x cos 45

=16.25 mm

change in y = 13.2 % of 16.25

= 2.14 .