# Example

A body moves in a simple harmonic motion and the following graph
gives the variation of its displacement x with time t.

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

b) Find the time duration in the first cycle
when the body is located above a displacement of 0.25 m.

[ x = x_{o} sin (1.05 t) , Δt =
1.9 s ]

## The
hint can be found in the model

## Solution:

a) x0 = 0.5 m

$\mathrm{since\; \omega}$ = 2 π T $\mathrm{,\; \omega}$ = 2 π 6.0$\mathrm{}$

ω = 1.05

x = x0sin(ω t) = 0.5 sin(1.05t)

b) At x = 0.25 m,

substituting back into the equation,

0.25 = 0.5 sin(1.05t)

$\mathrm{}$solving for t gives

0.5 = sin(1.05t)

$$

solving of the ranging of time, t

1.05t = 0.524 and π - 0.524

t_{1} =
0.5 , t_{2} = 2.4 s

looking at the graphical form of the
solution of the 2 times, t_{1} and
t_{2}.

giving the range to be t_{2} - t_{1} = 2.4 -
1.5 = 1.9 s

## Model:

http://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_SHM13/SHM13_Simulation.xhtml