# Math Help - Simple Harmonic Motion

1. ## Simple Harmonic Motion

In a small harbour, the depth of the water at high tide is 10m, and at low tide is exactly zero. The depth of water follows approximately simple harmonic motion with amplitude (A), 5m and time period (T), 12 hours. On one day, high tide is noon.

If d is the depth of the water above the mean depth (5m), write an equation of the form $d = A \ {\cos {\omega}t}$, substituting the appropriate values for $A$ and ${\omega}$.

Where, $t$ is time.

I already know $A = 5$ and ${\omega} = 0.5 \ rad \ h^{-1}$ or ${\omega} = 1.45 \ rad \ s^{-1}$ but i am not sure what the question is asking me to do. Does it want me to make a function for time so that when you input a time of day it will return an appropriate value? I am not sure, any ideas anyone?

Thanks

2. Originally Posted by steve@thecostins.co.uk
In a small harbour, the depth of the water at high tide is 10m, and at low tide is exactly zero. The depth of water follows approximately simple harmonic motion with amplitude (A), 5m and time period (T), 12 hours. On one day, high tide is noon.

If d is the depth of the water above the mean depth (5m), write an equation of the form $d = A \ {\cos {\omega}t}$, substituting the appropriate values for $A$ and ${\omega}$.

Where, $t$ is time.

I already know $A = 5$ and ${\omega} = 0.5 \ rad \ h^{-1}$ or ${\omega} = 1.45 \ rad \ s^{-1}$ but i am not sure what the question is asking me to do. Does it want me to make a function for time so that when you input a time of day it will return an appropriate value? I am not sure, any ideas anyone?

Thanks
they just want you to rewrite the formula with your values substituted.

how did you find $\omega$? perhaps i don't know how to deal with simple harmonic motion, but since the period is 12, i'd say $\omega = \frac {\pi}6$

so the answer using my values would be: $d = 5 \cos \frac {\pi t}6$

3. $\omega$ can be calculated with the equation ${\omega} = \frac{2\pi}T$

where $T$ is the period in time for one cycle or oscillation. In the case of the harbour, if working in hours, $T = 12$, or if working in seconds $T = 12 * 60 * 60$.

In the question why did they put "If d is the depth of the water above the mean depth"? Why mention of water ABOVE the mean depth?

So then i guess if they're looking for the answer in hours it would be:

$d = 5 \ \cos \frac{1}{2}t$

4. Originally Posted by steve@thecostins.co.uk
$\omega$ can be calculated with the equation ${\omega} = \frac{2\pi}T$

where $T$ is the period in time for one cycle or oscillation. In the case of the harbour, if working in hours, $T = 12$, or if working in seconds $T = 12 * 60 * 60$.

In the question why did they put "If d is the depth of the water above the mean depth"? Why mention of water ABOVE the mean depth?

So then i guess if they're looking for the answer in hours it would be:

$d = 5 \ \cos \frac{1}{2}t$
i still don't see how you got 1/2. if T = 12 then $\frac {2 \pi}{12} = \frac {\pi}6$ (and why did you change to seconds?)

it seems to me, we want to have a formula that calculates the depth of the water when it is above 5m. if that is the case, i think we need more information on what time of the day we are at and what time of the day do we have the max tide

5. Yeah, using $\frac{\pi}{6}$ is better than using $\frac{1}{2}$ because it's more accurate.

We know that a full cycle takes 12 hours, and the information given with the question says that high tide starts at noon.

6. Originally Posted by steve@thecostins.co.uk
Yeah, using $\frac{\pi}{6}$ is better than using $\frac{1}{2}$ because it's more accurate.

We know that a full cycle takes 12 hours, and the information given with the question says that high tide starts at noon.
yes, that is so. but what i am getting at is that we don't know where we are know. if we knew say, the height of the water was zero now, then we could judge and calculate for the height to be above the mean of 5m. but oh well. given the information we have, i think you have the best answer.

7. Originally Posted by steve@thecostins.co.uk
In a small harbour, the depth of the water at high tide is 10m, and at low tide is exactly zero. The depth of water follows approximately simple harmonic motion with amplitude (A), 5m and time period (T), 12 hours. On one day, high tide is noon.

If d is the depth of the water above the mean depth (5m), write an equation of the form $d = A \ {\cos {\omega}t}$, substituting the appropriate values for $A$ and ${\omega}$.

Where, $t$ is time.

I already know $A = 5$ and ${\omega} = 0.5 \ rad \ h^{-1}$ or ${\omega} = 1.45 \ rad \ s^{-1}$ but i am not sure what the question is asking me to do. Does it want me to make a function for time so that when you input a time of day it will return an appropriate value? I am not sure, any ideas anyone?

Thanks
"one day, high tide is noon"
So if the curve starts at a maximum, the sinusoidal function to use is cosine.

The general cosine equation is
y = A*cos(Bx -C) +D
where
A = amplitude----------------------------5m here
B = frequency; such that Period = 2Pi /B
C = horizontal shift
D = vertical shift
shift = relocation of the neutral axis from the x- or y-axis.

Since depth at high tide is given as = 10m and at low tide 0m, then the x-axis is at 0m. The mean depth, the neutral axis is then 5m above the x-axis. So the vertical shift is +5m.

If we start the curve at 12noon, at its maximum, then the horizontal shift is zero, and x (t in this Problem) starts at 12noon. Or t(0) is 12noon.

So the equation is
d = 5cos(B*t -0) +5 ----------(i)

period = 2pi/B
12hrs = 2pi/B
B = 2pi/12 = pi/6 cycle/hr

Hence,
d = 5cos*((pi/6)*t) +5
d = 5cos(pi*t / 6) +5 ------------------------the equation.

---------------------------------------------------
Check:

If the period is 12hrs, and the start is at 12noon where there is a high tide, then the nexy high tide is 12 hours from 12noon. That will be at 12midnight.
Then the low tide is at half of one period (in a cosine curve, the lowest point is at half of one period) which is at 6pm.

d = 5cos(pi*t /6) + 5

At 12noon, or t = 0,
d = 5cos(0) +5 = 5(1) +5 = 10m -----checks.

At 6pm, or t = 6 hrs,
d = 5cos(pi*6 /6) +5 = 5cos(pi) +5 = 5(-1) +5 = 0 -----checks.

At 12midnight, or t = 12 hrs,
d = 5cos(pi*12 /6) +5 = 5cos(2pi) +5 = 5(1) +5 = 10m -----checks.

8. Excellent Ticbol! I guess I never read the question carefully enough. I have one problem though
Originally Posted by ticbol
The general cosine equation is
y = A*cos(Bx -C) +D
where
A = amplitude----------------------------5m here
B = frequency; such that Period = 2Pi /B
C = horizontal shift
D = vertical shift
shift = relocation of the neutral axis from the x- or y-axis.
usually, the general equation is written as y = A*cos B(x - C) + D. here C is the horizontal shift. as you have it in your equation, C/B is the horizontal shift.

it didn't affect the answer here though, since C = 0, but it might in some other question. (or i don't know, maybe in practice it won't make much of a difference, since the value of C may be altered by the way you set up the equation or something)

Good Job though!

9. Originally Posted by Jhevon
Excellent Ticbol! I guess I never read the question carefully enough. I have one problem thoughusually, the general equation is written as y = A*cos B(x - C) + D. here C is the horizontal shift. as you have it in your equation, C/B is the horizontal shift.

it didn't affect the answer here though, since C = 0, but it might in some other question. (or i don't know, maybe in practice it won't make much of a difference, since the value of C may be altered by the way you set up the equation or something)

Good Job though!
I see.

B(x -C) = (Bx -BC)
I called "BC" as C

The frequency B remains the same.

Using the B(x -C) is sometimes confusing. Others interpret then that C is the phase angle or horizontal shift, when, in fact, it should be "C/B".
Confucing, is it not?
Should it not be "CB"?