Dimensional analysis. Waves on the surface of the ocean do not...

Question

Dimensional analysis. Waves on the surface of the ocean do not depend significantly on the properties of water such as density or surface tension. The primary 'return force' for water piled up in the wave crests is due to the gravitational attraction of the Earth. Thus the speed v (m/s) of ocean waves depends on the acceleration due to gravity g. It is reasonable to expect that υ might also depend on water depth h and the wave's wavelength λ. Assume the wave speed is given by the functional form v = Cgᵅ hᵝ λᵞ, where α , β , c and C are numbers without dimension.

(b) In shallow water, the speed of surface waves is found experimentally to be independent of the wavelength (i.e., γ = 0 in our assumed equation above for v). Using only dimensional analysis, determine the formula for the speed of waves in shallow water.

(b) In shallow water, the speed of surface waves is found experimentally to be independent of the wavelength (i.e., γ = 0 in our assumed equation above for v). Using only dimensional analysis, determine the formula for the speed of waves in shallow water.

Accepted Answer

Everyone in this problem, the relation we're given represents a physical quantity called drag force. So we have F is equal to C multiplied by row to the exponent alpha multiplied by B to the exponent beta multiplied by A, to the exponent gamma. Where alpha beta, gamma and C are dimensionless quantities. Row is the density of fluid V is the velocity and A is the cross sectional area. And we're asked to use dimensional analysis to determine the formula for the drag force. They were given four answer choices A through D and each of them have the form formula for F that we were given in the problem with different values for alpha beta and gamma. OK. We're gonna come back to these as we work through this problem. But our main goal is to figure out what those values of alpha, beta and gamma should be. All right. So we wanna use dimensional analysis. So let's start by thinking about the dimensions of the quantities we were given. So let's think about the force F right now forces. What are they measured in? Well, the typical unit we see forces in is newtons or call that a Newton is equal to a kilogram meter per second squared. So when we're talking about the dimensions of a force, we have a mass multiplied by a link divided by a time squared. OK? And again, in those standard units, kilogram meter per second squared. All right. So that's for F we're told C is dimensionless. So we don't need to worry about that. What about row? Well, row is the density, right? And recall that when we have a density, we take a mass and we divide it by a lawyer. OK. So what are the dimensions of a volume? How do we calculate volume or recall that a volume is going to be a length cubed? OK. So the dimensions of the density are gonna be mass divided by length cubed. And you can think of many different, you can think of a meters cube centimeters cube millimeters cube to measure a density or a volume. OK. That's where that denominator comes from. All right, moving along alpha is a dimensionless quantity. VV, we're told is a velocity. OK? And recall that a velocity is gonna be measured in a length her time. We're used to seeing meters per second. Meters measures length seconds measures time. So length divided by time. And finally, we have the dimensions of a, we're told that a is a cross sectional area for an area. We have the dimensions of length squared. All right. So we have the dimensions of all of the variables in this equation F row V and A and we know that the dimensions on the left hand side have to equal the dimensions on the right hand side for this equation to be consistent. OK. So let's substitute in everything we've just found and set them equal to each other. So on the left hand side, we have the dimensions of force mass multiplied by length divided by time squared. And on the right hand side C is dimensionless. So we don't need to include it in our dimensional analysis row has units of ma or dimension sorry, of mass divided by length cubes. And that's all raised to the exponent alpha. Then we have the velocity which has dimensions of length divided by time, all raised to the exponent beta. And finally, the area with dimensions L squared all raised to the exponent gamma, right, let's do some simplify. On the right hand side, we're gonna leave the left hand side alone mass multiplied by length multiplied by T times squared. We can apply these exponents to both the numerator and denominator. So we have M mass to the exponent alpha divided by a length to the exponent three alpha. And when we have an exponent raised to an exponent, we can multiply those exponents together. And we're gonna multiply it by a length to the exponent beta divided by time to the exponent beta multiplied by a lengthy exponent to gamma. A applying those same rules. Now, let's simplify and collect like terms. On the right hand side, we want a single M term, a single L term, a single T term so that we can compare the exponents to um from the left hand side. So on the left hand side, mass multiplied by a length divided by T squared, we're leaving that alone. On the right hand side, the only M term we have is M to the exponent alpha. OK. So we have M to the exponent alpha for the L terms we have L to the exponent beta, L to the exponent two gamma in the numerator and L to the exponent three alpha in the denominator. So we can simplify this recall that if we're multiplying two terms together with the same base, we can add exponents. And when we're dividing, we subtract exponents. So we're gonna get is L to the exponent beta plus two gamma minus three alpha. And then we're dividing by T to the exponent beta. All right. And now this is a matter of comparing. So on the left hand side, our M term just has an exponent of one, there's no exponent there, that indicates just an exponent of one. And so alpha is going to be equal to one, same thing. We have T squared in the denominator, right? That has an exponent of two. On the right hand side, we have T to the exponent beta in the denominator. And so beta must speak, what's it? Two? Mhm. Now, finally, we have L with an exponent of beta plus two gamma minus three alpha. On the left hand side, we have L with just an exponent of one. So we must have beta plus two gamma minus three alpha equal to one. And we can substitute in the value for beta. We just found too plus two gamma minus three multiplied by alpha, which we found to be one. This is all gonna be equal to one. OK. So simplifying two plus two, gamma minus three is equal to one. So two gamma minus one is equal to one, adding one to both sides, two gamma is equal to two. And finally dividing by two, we get that gamma is also equal to one. OK? So we found our alpha beta and gamma values that make this equation consistent that have the right dimensions. And so that is going to be the correct equation or the force. OK. So if we write out the equation now with these um exponents, we had that the force was equal to that constancy multiplied by row to the exponent alpha, which is just one. So we're just gonna leave it as row multiplied by V to the exponent beta, which we found to be two multiplied by A to the exponent gamma, which we found to just be one. And so our equation is gonna be F is equal to C multiplied by row multiplied by V squared, multiplied by A. Let's compare this to our answer choices. Yeah. Option A has exponents of two on the row and the A, that's not what we found. Option B has an exponent of three on A not what we found. Option C looks good. And option D only has exponent one all throughout. So that's also not correct. And so C is indeed the correct answer F is equal to C row V squared. A thanks everyone for watching. I hope this video helped you in the next one.

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