A buoy oscillates in simple harmonic motion as waves go past. At a given time it is noted that the buoy moves a total of 3.5 feet from its low point to its high point, and that it returns to its high point every 10 seconds. Write an equation that describes the motion of the vuoy of it is at its high point at time t=0
要有計算方法!!!
關於simple harmonic motion!! 15點!!
2008-12-12 7:11 pm
回答 (1)
2008-12-13 4:25 am
✔ 最佳答案
The equation of the wave is given by y = Asin(wt - kx)Amplitude, A =3.5 / 2 = 1.75
Angular frequency, w = 2 π / T = 2π / 10 = 0.628
k = 2π / λ
At t = 0, the buoy is at its highest point,
i.e. Asin(0 - kx) = A
sinkx = -1
kx = -π / 2
Hence, the equation of motion of the buoy is:
y = 1.75sin(0.628t - (-π / 2))
y = 1.75sin(0.628t + π/2)
Where distance is measured in feet.
t is measured in terms of seconds.
參考: Myself~~~
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