Lewis Harvey

2021-08-20

A bouy floating in the ocean is bobbing in simple harmonic motion with period 6 seconds and amplitude 3ft. Its dispfacement d from sea level at time $t=0$ seconds is 0 ft, and initially it moves downward. (Note that downward is the negative direction.)
Give the equation modeling the displacement d as a function of time t.

rogreenhoxa8

Step 1
The general equation of simple harmonic motion is given as,
$d=A\mathrm{sin}\left(\omega t+\varphi \right)$
Here,
d=Displacement from the mean position at any time t
A=Amplitude of the simple harmonic motion=3 ft.
$\omega$=The angular frequency of the motion
$\varphi$=The initial phase
Step 2
Using the relation, $\omega =\frac{2\pi }{T}$
T is the period of the simple harmonic motion
$\omega =\frac{2\pi }{6}rad/s$
Hence, the equation of simple harmonic motion becomes,

Assuming the direction above the surface as positive and the direction below the surface as negative, and the surface as the mean position.
At

$\mathrm{sin}\varphi =0$

Therefore, the equation of simple harmonic motion of the body is,

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