Shm (1 Viewer)

[redslert]

i cant do this i have no idea!!!!
massive headache help me out

A ship needs 11m of water to pass down a channel safely. At low tide, the channel is 6m deep and at high tide, 14m deep. Low tide is at 5.30am and high tide at 11.30am. Assume that the tidal motion is SHM.

(a) State the amplitude and the period of the motion
(b) Show that the earliest time that a ship can pass through the entrance is approx. 9am.

i got another one of these, but hopefully after someone showing me how to do this one i can do the other one! thank you very much!!!!!

 

[ND]

Similar kind of question: http://www.boredofstudies.org/community/showthread.php?s=&threadid=12899

 

[Giant Lobster]

Amplitude:

min = 6
max = 14
so logically the oscillations will go back and forth between 6 and 14 about the midpoint of 6 and 14. The amp. would then be (14-6)/2 = 4.

Period:

Low tide @ 5:30 (or 5.5)
High tide @ 11.30 (or 11.5)
using the symmetry of SHM, the period of motion can be found by (11.5-5.5)*2 = 12 (12 hrs per oscillation)

Part b:

with the obtained information u can find the equation of motion quite simply. I know there are many ways to do this, but the following way I find most easy:

Since the SHM starts at an extremity, i.e. low tide, let the motion be defined by x = -4cos(nt + a) where a is the phase.
We know it starts at 5.5 (5:30 am) so noting this, we can eliminate a by letting the origin of the number plane be (5.5,10). (i prefer to do this than to find a)

Now to find n: we know the period is 12 = 2*pi / n (by definition)
so n = pi/6

Thus x = -4cos(pi*t/6) with the origin at 5.5 (remember)
For 11m of depth, x needs to be 1 (because origin at (5.5,10))

so to solve 1 = -4cos(...) is quite simple. The first solution of t comes down to be 3.4826... but remembering the origin is at x = 5.5, u add 5.5 to this result, giving 8.9826... which is approx 9.

 

[jogloran]

My tutor says he hasn't seen a 'tide' SHM question in the 3 unit paper for a while. True?

 

[Giant Lobster]

regardless... these 'applied' shm questions r essentially all the same

if u do one, uve done em all basically.

 

[Jumbo Cactuar]

Originally posted by Giant Lobster

Period:

Low tide @ 5:30 (or 5.5)
High tide @ 11.30 (or 11.5)
using the symmetry of SHM, the period of motion can be found by (11.5-5.5)*2 = 12 (12 hrs per oscillation)

Although I hate SHM and am pretty bad at it, wouldn't one oscillation be the time taken to go from any one point to returning to that same point with the same v and a. Which is 24 hrs. Ok maybe I'm wrong,....

 

[Hotdog1]

yep. I asked a question like that some time ago. The period is a complete oscillation, the time from the maximum to the minimum is only half of the period.

 

[kazerati]

yes, i can understand ur confusion...
but, just from havin a fresh look at the Lobsters workings.. note that he did in fact multiply the difference between the maximum and the minimum by 2.. therefore a complete oscillation/period..
so that makes 12 hours correct