A bouy floating in the ocean is bobbing in simple harmonic motion with

Lewis Harvey

Lewis Harvey

Answered question

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.

Answer & Explanation

rogreenhoxa8

rogreenhoxa8

Skilled2021-08-21Added 109 answers

Step 1
The general equation of simple harmonic motion is given as,
d=Asin(ωt+ϕ)
Here,
d=Displacement from the mean position at any time t
A=Amplitude of the simple harmonic motion=3 ft.
ω=The angular frequency of the motion
ϕ=The initial phase
Step 2
Using the relation, ω=2πT
T is the period of the simple harmonic motion
ω=2π6rad/s
Hence, the equation of simple harmonic motion becomes,
d=(3 ft.)sin(2π6t+ϕ)
Assuming the direction above the surface as positive and the direction below the surface as negative, and the surface as the mean position.
At t=0 sec. d=0
(3 ft.)sin(2π6(0)+ϕ)=0
sinϕ=0
ϕ=π radians
Therefore, the equation of simple harmonic motion of the body is, d=(3 ft.)sin(π3t+π)

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