A ship needs 7.5m of water to pass down a channel safely. At high tide the channel is 9m deep and at low tide the channel is 3m deep. High tide is at 4:00am, Low tide is at 10:20am. Aussume that the tide rises and falls in simple harmonic motion
What is the latest time before noon, to the nearest minute that the ship can safely proceed through the channel?
answer is 6:06 am
1 Answer | Add Yours
Assuming simple harmonic motion, we can model with a sinusoid of the form `y=Asin(B(x-k))+h` .
Here A is the amplitude, B comes from the period, h the horizontal translation and k the vertical translation.
A: The maximum is 9 and the minimum is 3 so the amplitude is found by `A=(9-3)/2=3`
B: The period is 760 minutes. (There are 380 minutes from high tide to low tide.) Then `B=(2pi)/p=(2pi)/760=pi/380`
h: The sine is zero at x=0; the model hits the midline at x=50 so h=50. (There are 380 minutes from high tide to low tide -- so 190 minutes from either extreme the tide is at the midline. Subtracting 190 from 240 yields the 50 minute mark.)
k: The vertical translation describes the midline and `k=(9+3)/2=6`
Putting it all together the model is: `y=3sin(pi/380(x-50))+6`
We want to know when y=7.5:
`.5=sin(pi/380(x-50))` Taking the arcsin of both sides:
This is in minutes; to convert to hours we divide by 60:
`x("hours")=17/9~~1:53` or 1hour 53 1/3 minutes.
However, since we used the arcsin, we need to check the other possible solutions. Instead of `pi/6` we could use `(5pi)/6` .
Then we get :
`x=1100/3` and converting to hours `x=6 1/9` hours. Since 1/9 of an hour is `6 2/3` minutes we get 6:06am. (By 6:07 the depth is below 7.5m)
The graph: x in units of 190 minutes
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes