2004 tsunami? how to calcualte?

[ritva_1950]

calculate 2004 tsunami by the following equations to calculate the velocity(c), wavelength (l) and amplitude (h) of the tsunami for water depths (d) of 5 km (abyssalplain), 1.5 km (mid-ocean ridge) and 10 m (near shore).
Why did seawater recedefrom the shore prior to the arrival of the tsunami in Thailand but not in SriLanka?Equations:
c = (d.g)^1/2
l µ c
h µ 1/c^1/2
g = 10 m.s^-2





µ = what mean this sign?

 

[wjm11]

calculate 2004 tsunami by the following equations to calculate the velocity(c), wavelength (l) and amplitude (h) of the tsunami for water depths (d) of 5 km (abyssalplain), 1.5 km (mid-ocean ridge) and 10 m (near shore).
Why did seawater recedefrom the shore prior to the arrival of the tsunami in Thailand but not in SriLanka?Equations:
c = (d.g)^1/2
lµ c
h µ 1/c^1/2
g = 10 m.s^-2

µ = what mean this sign?

The way you are using "\(\displaystyle \mu\)" it means "is proportional to". For example, \(\displaystyle h\, \mu\, (1/c)^{1/2}\) means the amplitude of the tsunami wave is proportional to the square root of 1/c.

You can find a nice explanation here:

http://terrytao.wordpress.com/2011/03/13/the-shallow-water-wave-equation-and-tsunami-propagation/

 

[ritva_1950]

The way you are using "µ " it means "is proportional to". For example, h µ (1/c)^1/2 means the amplitude of the tsunami wave is proportional to the square root of 1/c.

You can find a nice explanation here:

http://terrytao.wordpress.com/2011/03/13/the-shallow-water-wave-equation-and-tsunami-propagation/

so for 5km (5000 m) mulitply 10 is 50000 than square root around 224 m/s that is C and Y ok!
and 1/224 is 0.00044 and square root of 0.0044 is 0,02 so 0.02 for h!

is everything cooorect????????????????

 

[wjm11]

so for 5km mulitply 10 is 50000 than square root around 224 m/s that is C and Y ok!

and 1/224 is 0.00044 and square root of 0.0044 is 0,02 so 0.02 for h!

is everything cooorect????????????????

I do not know what "and Y" means. Your velocity seems okay.

1/224 = .000446

sqrt(.000446) = .067m

However, this answer seems too small. The above referenced site provides an example:

"... a tsunami with initial amplitude of one metre at a depth of 2 kilometres..."

Another example is provided here:

http://www.maths.qmul.ac.uk/~jel/MTH6129/OWP/notes/MTH6129part4c.pdf

and offers 1 m amplitude with a depth of 4 km as an example.

 

[ritva_1950]

I do not know what "and Y" means. Your velocity seems okay.

1/224 = .000446

sqrt(.000446) = .067m

However, this answer seems too small. The above referenced site provides an example:

"... a tsunami with initial amplitude of one metre at a depth of 2 kilometres..."

Another example is provided here:

http://www.maths.qmul.ac.uk/~jel/MTH6129/OWP/notes/MTH6129part4c.pdf

and offers 1 m amplitude with a depth of 4 km as an example.

hallo again!
thxs for a nice copeartion!

i mean lµ c = both l and c is the same , also same answear for them both??? or i get i wrong ??????????
like if you know C you know l too!?????????????

 

[wjm11]

hallo again!
thxs for a nice copeartion!

i mean lµ c = both l and c is the same , also same answear for them both??? or i get i wrong ??????????
like if you know C you know l too!?????????????

No, l and c are not the same. "Proportional to" does not mean "equal to". You would need to know the proportionality constant to calculate l if you already know c. Remember that in your problem, l is wavelength and c is velocity. Obviously, these are not the same thing. To say that wavelength and velocity are proportional means that if you double the velocity, then the wavelength will be doubled, too.

 

[floringaspar19]

Hi

I have the same exercise to do and I was wondering if you've done this exercise please paste the answers on my email adress: florin.gaspar@hotmail.com. Thanks, Florin