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### 1.2.7 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

### 1.2.7.1 Solution:

a) x0 = 0.5 m

$\mathrm{since\; \omega}=\genfrac{}{}{0.1ex}{}{2\pi}{T}$ $\mathrm{,\; \omega}=\genfrac{}{}{0.1ex}{}{2\pi}{6.0}$$\mathrm{}$

ω = 1.05x = 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

### 1.2.7.2 Model:

### For Teachers

### Translations

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