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Chapter 11: Problem 69

In contrast to deep-water waves, shallow ripples on the surface of a pond are due to surface tension. The surface tension \(\gamma\) of water characterizes the restoring force; the mass density \(\rho\) of water characterizes the water's inertia. Use dimensional analysis to determine whether the surface waves are dispersive (the wave speed depends on the wavelength) or non dispersive (their wave speed is independent of wavelength). [Hint: Start by assuming that the wave speed is determined by \(\gamma, \rho,\) and the wavelength \(\lambda .1]\)

Short Answer

Expert verified
Answer: Dispersive

Step by step solution

01

Identify the physical quantities and their dimensions

We have the following physical quantities: - Surface tension (\(\gamma\)) with dimensions of force per length [M^1 L^0 T^-2 L^-1] - Mass density (\(\rho\)) with dimensions of mass per volume [M^1 L^-3 T^0] - Wavelength (\(\lambda\)) with dimensions of length [M^0 L^1 T^0] - Wave speed (\(v\)) which we need to determine, with dimensions [M^0 L^1 T^-1]
02

Use the Buckingham Pi theorem for dimensional analysis

The Buckingham Pi theorem states that there must be a dimensionless product involving the given variables. Let's consider a product involving these variables and the wave speed (\(v\)): \(\Pi = \gamma^a \rho^b v^c \lambda^d\) Now, we need to find the exponents (\(a, b, c, d\)) such that the product is dimensionless.
03

Set up the dimension equation and solve for exponents

To make the product dimensionless, the dimensions of all the involved variables should cancel out. Let's set up the dimension equation: \([M^1 L^0 T^{-2} L^{-1}]^a [M^1 L^{-3} T^0]^b [M^0 L^1 T^{-1}]^c [M^0 L^1 T^0]^d = [M^0 L^0 T^0]\) Now, equate the powers of Mass (M), Length (L), and Time (T) on both sides: - For Mass (M): \(a + b = 0\) - For Length (L): \(-a - 3b + c + d = 0\) - For Time (T): \(-2a - c = 0\)
04

Solve the system of equations for exponents

Let's solve the system of equations we got in Step 3: 1. From the equation for Mass (M), we get \(b = -a\) 2. Substituting \(b = -a\) in the equation for Length (L), we get \(-a + 3a + c + d = 0 \implies a + c + d = 0\) 3. Substituting \(b = -a\) in the equation for Time (T), we get \(-2a - c = 0 \implies c = -2a\) Finally, substitute \(c = -2a\) in the equation from point 2, we get: \(a - 2a + d = 0 \implies d = a\)
05

Construct the dimensionless product and determine wave speed dependence

From Step 4, we have found the exponents: \(a, b, c, d\) \(b = -a\), \(c = -2a\), and \(d = a\) Let's use these values in our dimensionless product \(\Pi\): \(\Pi = \gamma^a \rho^{-a} v^{-2a} \lambda^a\) Now, we rewrite the equation by isolating the wave speed: \(v^{-2a} = \frac{\gamma^a}{\rho^a \lambda^a}\) Take the power of \(1/2a\) on both sides to get the wave speed: \(v = \left(\frac{\gamma}{\rho \lambda}\right)^{\frac{1}{2}}\)
06

Analyze the dispersion property of the wave

From the expression for the wave speed, we can see that it depends on the wavelength, \(\lambda\). This means that the waves are dispersive since the wave speed is not constant for different wavelengths. Wave speed equation: \(v = \left(\frac{\gamma}{\rho \lambda}\right)^{\frac{1}{2}}\) Hence, the surface waves on a pond are dispersive as the wave speed depends on the wavelength.

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Most popular questions from this chapter

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