Harmonic Oscillator 4

1. Jun 4, 2014

Karol

1. The problem statement, all variables and given/known data
In a port the tide and the low tide change with harmonic motion. at the high tide the water level is 12 meters and at the low tide it is 2 meters. between the tide and the low tide there are 6 hours.
A ship needs 8 meters of water depth. how long can it stay in the port

2. Relevant equations
Harmonic motion: $$x=A\cos(\omega t)$$
The period: $$T=\frac{2\pi}{\omega}$$

3. The attempt at a solution
The difference in depth between tide and low tide is 10 meters, so the middle point, the 0 point is at 7 meters. the ship can stay until the water reaches water level of 8 meters which is 1 meter above the 0 level.
$T=\frac{2\pi}{\omega}\rightarrow 12\times 3600=\frac{2\pi}{\omega}\Rightarrow \omega=0.000145$
$x=A\cos(\omega t)\rightarrow 1=5 \cos(0.000145\cdot t)\Rightarrow 0.000145\cdot t=78.5^0$
$\rightarrow 1.37[rad]=0.000145\cdot t \Rightarrow t=9444[sec]=2.6[hour]$
It should be 5.2 hours, although i can't understand why since it is almost the time between tide and low tide and we need a point above the 0 point

Last edited: Jun 4, 2014
2. Jun 4, 2014

tms

Unless there is a typo in the question, the high tide is 12 m and the low tide 6 m, so the difference is not 10 m.

Also, it always helps to solve problems symbolically, only plugging in numbers at the very last step. That way you get to see what is going on in the physics; when you plug in numbers you lose information.

3. Jun 4, 2014

Karol

Right, my mistake, the low tide is 2 meters. i fixed it.

4. Jun 4, 2014

CAF123

I get the same answer as you. The wave can be built from a shifted cosine wave $y = 5\cos(\omega t) + 7$, where $\omega = 2\pi/T$. Sub in T = 12 gives $\omega = \pi/6 hr^{-1}$.

When y = 8, we find t ≈ 2.6 hours. So if the ship docked at the port when the tide was 12m high, he could stay for this amount of time until the water level went down to 8m.

5. Jun 4, 2014

tms

The question is asking for the time between when the tide rises through 8 m and when it falls through 8 m, so your answer should be doubled.

6. Jun 4, 2014

CAF123

I see, I misinterpreted it. So for t in [0,2.6] and t in [9.4,12] the ship is okay. This amounts to 5.2 hrs indeed.