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Need Guidance: IsaacPhysics - Safe Passage Question

https://isaacphysics.org/questions/safe_passage?board=bfl22_shm_quantities_calc

Hello everyone,
I am stuck with this question (link attached above).

I tried drawing the graph (x-axis is time in seconds and y-axis is depth in metres) and got an equation for it.
I set it equal to 9.0m but I keep getting the answer wrong.

The equation I got is: y = 8.5 - 2.5cos(((2 * pi) / 45600) * x - 43200)

How I got the equation:
1. I modelled the graph as a cos curve and used the equation from the equation sheet: displacement = amplitude * cos(angular speed * time).
2. It's amplitude is 2.5, which is 2.5*cos((2*pi) / (2(6*3600 + 20*60))*t) = 2.5cos(((2*pi)/(45600))*t).
3. But it is wrong so far, as it is reflected in the x-axis, so it should be -2.5cos(((2*pi)/(45600))*t).
4. It's been shifted up by 8.5, so 8.5 - 2.5cos(((2*pi)/(45600))*t).
5. It has been shifted right by 12hrs, so take away (12*3600=) 43200 from inside the cos part.

Please may I know if my equation is wrong or if my approach is wrong.

Thank you all for your help.
Reply 1
Looks like youre about right, but overcomplicating it a bit by working in seconds (the question asks for minutes) and you may as well take noon to the the origin and forget about it in the formula so
8.5 - 2.5*trig(...)
fits the bill and choosing trig() as cos() is the simplest thing as its 6 when cos is 1 or the arg is 0 and 11 when cos is -1 or arg is pi. So you want the angular speed in rad/min so
wT = 2pi
and T is the time period in mins so ... then its just
8.5 - 2.5*cos(wt) = 9
where t is the time in minutes from noon ...
(edited 2 months ago)
Reply 2
Original post by mqb2766
Looks like youre about right, but overcomplicating it a bit by working in seconds (the question asks for minutes) and you may as well take noon to the the origin and forget about it in the formula so
8.5 - 2.5*trig(...)
fits the bill and choosing trig() as cos() is the simplest thing as its 6 when cos is 1 or the arg is 0 and 11 when cos is -1 or arg is pi. So you want the angular speed in rad/min so
wT = 2pi
and T is the time period in mins so ... then its just
8.5 - 2.5*cos(wt) = 9
where t is the time in minutes from noon ...

Thank you so much for helping me. (I had spent nearly 9 hours on this question trying different things.) I used your approach and I managed to find the correct answer. Thank you so much.
(edited 2 months ago)
Reply 3
Original post by azazazaz91345
Thank you so much for helping me. (I had spent nearly 9 hours on this question trying different things.) I used your approach and I managed to find the correct answer. Thank you so much.

You were not far off, just think about units, origin, .... and pick stuff that makes it easier.

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