# Example

A particle moves in a simple harmonic motion between O to A and B
and back to O as shown. The positions B, B1, O, A1 and A are
equally spaced. The time taken to travel from A to B is 5.00 s and
the distance AB is 4.00 m.

a) Write an equation to represent the given
simple harmonic motion.

b) During one cycle, find the time that the
particle stays in the region:

i) A-A1

ii) A1-B1

[ x = 0.200
sin (0.628 t) , 0.833 s , 3.33 s ]

## Hint:

## Solution:

a) since x0 is
half of AB = 4 m

x0 = 2.00 m

the equation
is x = x0 sin(ωt)

to find ω, and
we can given T, we use

$\omega $ = 2 π T = 2 π ( 5 ) ( 2 ) = 0.628

thus the
equation is x = 2.00 sin(0.628t)

b) the
strategy is to find the time which all the points B, B1, O, A1 and
A occurs.

to find time
from A to A1,

when x = A1 =
1.00, imply 1.00 = 2.00 sin (0.628t), therefore t_{A1} = 0.833 s

similarly,
when x = A = 2.00, imply 2.00 = 2.00sin (0.628t), therefore t_{A} = 2.50 s

time from A1
to A = t_{A} -t_{A1} = 2.50 - 0.833 = 1.66
s

therefore time
from A1 to A to A1 = (1.66)(2) = 3.33 s

b) since we
know t 0 to A1 = 0.833 s

by comparing
the 4 equal sections of OA1 = A1O = OB1 = B1O = 0.833 s

the total time
spent inside A1B1 = (4)(0.833) = 3.33 s

## Model

http://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHM14/SHM14_Simulation.xhtml