Simple Harmonic Motion (1 Viewer)

[goobi]

Low tide at the mouth of a river occurs at 4pm and high tide at 10pm. Low tide is 4 metres and high tide is 12 metres. A ship needs 10 metres of water to pass safely through the mouth of the river. If the motion of the tides is take to be simple harmonic, what will be the first times during which the ship may pass?

What I've done:

Period = (10-4)*2=12*EDIT*
Therefore, n=pi/6
Equilibrium position = (4+12)/2 = 8m
Amplitude = 8-4 = 4m

And I don't know what to do after that...

Thanks in advance for any help

 

[fortune_cookie]

Note: you calculated the period incorrectly in your working. It should be 12hours. However, you got the value of n correct.

x = 8 -4cos[ (pi/6)t ]

You have to use the cos function because it starts at it's maximum value. Also, you subtract it because you want it to be initially at low tide.

10 = 8-4cos[ (pi/6) t ]

cos( (pi/6)t ) = -1/2

Related angles = cos^(-1) (1/2) = (pi/3)

Cos is negative in 2nd and 3rd quadrant

(pi/6)t = (pi- (pi/3)) , (pi + (pi/3) )

(pi/6)t = 2pi/3 , 4pi/3

t = 4, 8

Therefore first time = 4pm + 4hours = 8pm

 

[goobi]

Note: you calculated the period incorrectly in your working. It should be 12hours. However, you got the value of n correct.

x = 8 -4cos[ (pi/6)t ]

You have to use the cos function because it starts at it's maximum value. Also, you subtract it because you want it to be initially at low tide.

10 = 8-4cos[ (pi/6) t ]

cos( (pi/6)t ) = -1/2

Related angles = cos^(-1) (1/2) = (pi/3)

Cos is negative in 2nd and 3rd quadrant

(pi/6)t = (pi- (pi/3)) , (pi + (pi/3) )

(pi/6)t = 2pi/3 , 4pi/3

t = 4, 8

Therefore first time = 4pm + 4hours = 8pm

Thanks mate!

But I have two questions now. I wonder why the displacement function is x = 8 -4cos[ (pi/6)t ] but not 8 -4cos[ (pi/6)t + α] where α is a constant. Is the constant α necessary there? I am asking this because the textbook defines the displacement of a general particle is x= acos( nt + α)...

Also, why do we need to add 8 to -4cos[ (pi/6) t ] as opposed to only x = -4cos[ (pi/6)t ]? Is it because the equilibrium position is at x=8 ?

 

[RishBonjour]

Thanks mate!

But I have two questions now. I wonder why the displacement function is x = 8 -4cos[ (pi/6)t ] but not 8 -4cos[ (pi/6)t + α] where α is a constant. Is the constant α necessary there? I am asking this because the textbook defines the displacement of a general particle is x= acos( nt + α)...

Also, why do we need to add 8 to -4cos[ (pi/6) t ] as opposed to only x = -4cos[ (pi/6)t ]? Is it because the equilibrium position is at x=8 ?

about the plus a. the wave starts from LOW tide, so there is no SHIFT, its simply the cosine curve, upside down

and the second part, its because equilibrium position is at x=8

(I think)

edit:
also, you could express the same function as a sine function by ADDING a (since sine starts from equilibrium position, but here its from low tide)
BUT we look for the most appropriate function ===> cosine in this case