Find values of p & q in v=K(lambda^p)(g^q)

[glassy]

Homework Statement

The speed of ocean waves depends on their wavelength lambda (measured in meters) and the gravitational field strength g (measured in m/s^2) in this way:
v=K(lambda^p)(g^q)
where K is a dimensionless constant. Find the values of the exponents p and q.

Homework Equations

The Attempt at a Solution

P=(ln(v/k)-qlng)/(lnlambda)
q=(ln(v/k)/(lambda^p))/(lng)

 

[PeterO]

Homework Statement

The speed of ocean waves depends on their wavelength lambda (measured in meters) and the gravitational field strength g (measured in m/s^2) in this way:
v=K(lambda^p)(g^q)
where K is a dimensionless constant. Find the values of the exponents p and q.

Homework Equations

The Attempt at a Solution

P=(ln(v/k)-qlng)/(lnlambda)
q=(ln(v/k)/(lambda^p))/(lng)

this is a units or dimension question.

Suppose p was 3 and q was 5

then V = K.L^3.g^5 [it was shorter to type L than spell out lambda]

Look at the units.
L^3 would have units m^3
g^5 would have units m^5.s^-10

combines that makes m^8.s^-10

But this is velocity! so the units would be m.s^-1 so 3&5 are certainly not the values.

What should they be?

 

[glassy]

Am I looking for actual numbers?
The way I understand it, is that how quickly (how many meters per second) the crest of a wave moves across a distance depends on how far apart the crests are (in meters), then multiplied by the gravitational pull. The gravitational pull is measured in meters/second^2. I don't really understand the idea of gravitational pull being measured in meters per second^2. Also I don't understand why wavelength and gravitational pull are exponentiated.
You say that this is a dimensions or units question, which means I need to convert from one unit to another I guess, but I don't see what it is I am converting. We just have meters, and meters over seconds, and the answer is supposed to be meters over seconds, so what is the problem?

Thank you for your help thus far,
-A

 

[PeterO]

Am I looking for actual numbers?
The way I understand it, is that how quickly (how many meters per second) the crest of a wave moves across a distance depends on how far apart the crests are (in meters), then multiplied by the gravitational pull. The gravitational pull is measured in meters/second^2. I don't really understand the idea of gravitational pull being measured in meters per second^2. Also I don't understand why wavelength and gravitational pull are exponentiated.
You say that this is a dimensions or units question, which means I need to convert from one unit to another I guess, but I don't see what it is I am converting. We just have meters, and meters over seconds, and the answer is supposed to be meters over seconds, so what is the problem?

Thank you for your help thus far,
-A

Acceleration units can be confusing, because of the way we just bunch similar units together - that is why the m/s^2. For understanding it is not metres per second squared, even though it is written like that.

If you used a stop watch in a car, you might measure that it reached 60 km/h in 5 seconds.
That represents an average acceleration of 12 (km/h)/s or 12 kilometres per hour per second

I will use a comma to show where we pause when reading that:

Kilometres per hour, per second.
Seems clear but if we pause in the wrong place we would say

kilometres, per hour per second which sound like gobbledy-gook.

The real problem arises if the cars speedometer was calibrates in metres per second instead of kilometres per hour.
You might measure that the car accelerated to 15 m/s in 5 seconds.

That would mean an acceleration averaging 3 m/s each second - or 3 metres per second per second.
When written in english, with the comma [pause] for emphasis that is

3 metres per second, per second

However in symbols we write that m/s^2 the exponent indicating per second occurred twice.

If you multiply wavelength [m] by g [m/s^2] we get dimension/units of m^2/s^2, which can be written as (m/s)^2
That means lambda x g gives the square of speed.
to get just speed, we need the square root of that - and square root is shown as ^0.5

eg 9^0.5 = 3 [or -3]

so if the exponents in the original formula were both 0.5, the dimensions would be fine.