phy SHM - Ans.fyi 參考答案

phy SHM

用戶107913

2011-07-03 7:42 am

In a dock, the water depth varies between 30m and 10m
as the tide is at this highest and its lowest respectively
On a certain day, low tide occurs at 12:00 noon, and a high tide will occur at 5:45pm.

So ,from the given information, I can find the Period (T)
T= 41400s

but now, I want to find the time when the water depth is 20m

i tried this method to work out this answer,
x=Acoswt
10=10cos (2pi / 41400) t
but i can't find the answer

I know the answer maybe T/4
But how to work it out by using the Equation ?

回答 (1)

天同

2011-07-03 8:12 am

✔ 最佳答案

This equation is wrong: 10=10cos (2pi / 41400) t

Be aware that, in general, x is the displacement of the oscillating object measured from the equilibrium position. But you have used the displacement from the extreme position (i.e. the lowest position in this example). That's why you cannot get the right naswer.

When the water depth is 20 m, it is right at the equilibrium position (or mean position). Hence, x = 0 m. Using the equation you have quoted,

0 = A.cos(wt)
where A is the amplitude
i.e. cos(wt) = 0
or wt = pi/2 [just use the smallest value, instead of the general solution of pi(n+1/2) ]

But w = 2.pi/T, where T is the period
thus, (2.pi/T)t = pi/2
t = T/4

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