M3 question

The rise and fall of the water level in a harbour is modelled as simple harmonic motion. On a particular day the maximum and minimun depths of the water in the harbour are 10m and 4m these occur at 100 hours and 1700 hours respectively.

Find the speed in mh^-1 at which the water level is falling at 1600 hours on this particular day.

Find the total time between 1100 hours and 2300 hours on this particular day for which the depth of water in the harbour is less than 5.5m

Help appreciated
Rep for answers

Reply 1

Hanratarsed

Okay i hope this is right...i have a tendancy to make stupid mistakes.

If the rise and fall of the tide is modelled as SHM, with maximum 10m and minimum 4m, it has an amplitude of half this distance around the midpoint. So, we have SHM of amplitude 3m around the 7m mark.

As the water moves between the maximum and the minimum between 0100 and 1700, the time it takes to complete one period (from the maximum back to the maximum) is 16*2 hours = 32 hours. The angular velocity is given by 2*PI/T, and so the angular velocity is PI/16

As the motion starts at the top of its pattern, it can be modelled using the equation:

x = a * Cos wt

where a is amplitude, w is angular velocity and t is the time after 0100 hours

Differentiating this gives v = -awSinwt, and filling in the numbers gives 0.115 m/s.

The first equation can be rearranged to give the time at which the water level is at a specific height, as t = (Cos-1 (x/a))/w. By working out the times at which the water is at 5.5m (x= -1.5, as the water level is 1.5m below the midpoint) the final answer can be calculated.

I get T1 = 10:40 hrs, so occurs at 1140
And T2 = 21:20 hrs, so occurs at 2220

These times fit within the limits, and the gap between them is 10:40 hrs.

I hope that this helps and that I've got the right answer!

Luke

Reply 2

Enixa

I got stuck on this question too and I know it's been a while but I'll show the working in case anyone else gets stuck whilst revising for M3... The above solution is an alternative correct method but since you've written the times wrong, it's not correct- it's 1100 hours for the minimum and 1700 hours for the maximum.
So it's basically the same but the amplitude is 3 hours instead with a centre of oscillation at 1400 hours. The time period is this 2*6 (6 is from 1700-1100 hours) which is 12 hours and so the angular velocity 'w' is 2Pi/12 which is Pi/6

use the equation x=acoswt
so x= 3cos(pi/6)t
t= 1600-1100 so is 5 hours
x= 3cos(pi/6)*5
so x= 3root3/2

then using the equation
v^2=w^2(a^2-x^2)
v^2= (Pi/6)^2(9-27/4)
and then just work through to find v...

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