The figure below shows the displacement of a simple harmonic oscillator as a fun
ID: 1306992 • Letter: T
Question
The figure below shows the displacement of a simple harmonic oscillator as a function of time. (The grid spacing on the t-axis is 4 s and on the y-axis is 10 cm.)
(a) Find the period, frequency, and amplitude of the oscillator.
(b) At what time(s) does the acceleration have its largest positive value? (Select all that apply.)
4 s
8 s
12 s
16 s
20 s
24 s
28 s
32 s
(c) When does the acceleration have its most negative value? (Select all that apply.)
4 s
8 s
12 s
16 s
20 s
24 s
28 s
32 s
period = s frequency = Hz amplitude = cmExplanation / Answer
a) period is the time for one full cycle. count boxes along the x-axis.
if period is the time (in s) required for one full cycle, then frequency is the inverse of the period - i.e. the number of full cycles per second. amplitude is the absolute value of the maximum value on the y-axis. there is a max pos value and a max neg value. the max pos value is the same as the absolute value of the max neg value. that's the amplitude.
So, period = 16
Frequency = 1/16 Hz
Amplitude = 20cm
b) and c) the acceleration is maximal at the times where the displacement is maximal. acceleration is zero when the amplitude is zero, and at that point the speed is the greatest.
i.e. there are 2 times where there is maximum acceleration during one cycle. one at the max pos displacement (max neg acceleration), and the other at the max neg displacement (max pos acceleration).
So,max acceleration at (b) 12s,28s
min acceleration at (c) 4s,20s