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

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In summary, the speed of ocean waves depends on their wavelength lambda and the gravitational field strength g. The values of the exponents p and q are found by solving the equation v=K(lambda^p)(g^q).
  • #1
glassy
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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)
 
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  • #2
glassy said:

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?
 
  • #3
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
 
  • #4
glassy said:
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.
 
Last edited:
  • #5


I would approach this problem by first understanding the physical relationship between the variables involved. The equation v=K(lambda^p)(g^q) suggests that the speed of ocean waves (v) is directly proportional to both the wavelength (lambda) and the gravitational field strength (g), with the constant K representing the relationship between these variables.

To find the values of the exponents p and q, I would begin by rearranging the equation to isolate each variable. This would give me:
ln(v/K) = p*ln(lambda) + q*ln(g)

From this, I can see that the exponents p and q represent the degree to which the wavelength and gravitational field strength affect the speed of ocean waves, respectively. To find their values, I would take the natural logarithm of both sides of the equation and use basic algebra to solve for p and q. The resulting values would depend on the specific values of the constant K and the speed of ocean waves (v) that are given in the problem.

Additionally, I would also consider the units of the variables involved and make sure that the resulting values for p and q have appropriate units. For example, since wavelength is measured in meters and gravitational field strength is measured in m/s^2, the units for p and q should cancel out to give a unit of meters per second for the speed of ocean waves (v).

In conclusion, the values of p and q in the given equation depend on the physical relationship between the variables and can be found by manipulating the equation and considering the units of the variables.
 

1. What is the equation v=K(lambda^p)(g^q) used for?

This equation is used in physics to calculate the velocity (v) of an object based on the wavelength (lambda) and gravitational acceleration (g), with the constants K, p, and q representing other relevant factors.

2. How do you find the values of p and q in this equation?

To find the values of p and q, you will need to have values for v, lambda, and g, as well as the constant K. Then, you can rearrange the equation to solve for p and q using algebraic manipulation.

3. What do the variables lambda and g represent in this equation?

The variable lambda represents the wavelength of the object, which is the distance between two consecutive peaks or troughs of a wave. The variable g represents the gravitational acceleration, which is a constant value that determines the force of gravity on an object.

4. What is the significance of the constant K in this equation?

The constant K represents any additional factors or forces that may affect the velocity of the object besides wavelength and gravitational acceleration. It is often used to account for external factors such as air resistance or friction.

5. Can this equation be used for any type of object or situation?

Yes, this equation can be used for any object or situation where the velocity is dependent on wavelength and gravitational acceleration. However, it may need to be modified or adjusted for specific scenarios, such as objects moving through different mediums or in different gravitational environments.

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