To determine the dependency of the speed of water surface waves (capillary waves) on the density of water (ρ), their wavelength (λ), and surface tension (γ), we can use dimensional analysis. Here’s a step-by-step breakdown of the solution:
### Step 1: Identify the Dimensions
- **Density (ρ)**: The dimension of density is mass per unit volume, which is given by:
\[
[\rho] = M L^{-3}
\]
- **Surface Tension (γ)**: Surface tension is defined as force per unit length. The dimension of force is mass times acceleration (F = ma), which gives:
\[
[\text{Force}] = M L T^{-2}
\]
Therefore, the dimension of surface tension is:
\[
[\gamma] = \frac{M L T^{-2}}{L} = M T^{-2}
\]
- **Wavelength (λ)**: The dimension of wavelength is simply length:
\[
[\lambda] = L
\]
- **Speed (v)**: The dimension of speed is length per unit time:
\[
[v] = L T^{-1}
\]
### Step 2: Set Up the Dimensional Equation
Assume that the speed of the water surface waves (v) can be expressed as a function of density (ρ), surface tension (γ), and wavelength (λ):
\[
v = k \cdot \rho^a \cdot \gamma^b \cdot \lambda^c
\]
where \( k \) is a dimensionless constant, and \( a \), \( b \), and \( c \) are the powers we need to determine.
### Step 3: Write the Dimensions in Terms of a, b, and c
Substituting the dimensions into the equation gives:
\[
[L T^{-1}] = [\rho]^a \cdot [\gamma]^b \cdot [\lambda]^c
\]
This can be expressed as:
\[
L T^{-1} = (M L^{-3})^a \cdot (M T^{-2})^b \cdot (L)^c
\]
Expanding this, we have:
\[
L T^{-1} = M^{a+b} L^{-3a + c} T^{-2b}
\]
### Step 4: Equate the Powers of Each Dimension
Now, we equate the powers of \( M \), \( L \), and \( T \) from both sides of the equation:
1. For mass (M):
\[
a + b = 0 \quad \text{(1)}
\]
2. For length (L):
\[
-3a + c = 1 \quad \text{(2)}
\]
3. For time (T):
\[
-2b = -1 \quad \text{(3)}
\]
### Step 5: Solve the Equations
From equation (3):
\[
b = \frac{1}{2}
\]
Substituting \( b \) into equation (1):
\[
a + \frac{1}{2} = 0 \implies a = -\frac{1}{2}
\]
Substituting \( a \) into equation (2):
\[
-3(-\frac{1}{2}) + c = 1 \implies \frac{3}{2} + c = 1 \implies c = 1 - \frac{3}{2} = -\frac{1}{2}
\]
### Step 6: Write the Final Expression
Now substituting \( a \), \( b \), and \( c \) back into the equation for speed:
\[
v = k \cdot \rho^{-\frac{1}{2}} \cdot \gamma^{\frac{1}{2}} \cdot \lambda^{-\frac{1}{2}}
\]
This can be rearranged to:
\[
v = k \cdot \sqrt{\frac{\gamma}{\rho}} \cdot \frac{1}{\sqrt{\lambda}}
\]
### Conclusion
Thus, the dependency of the speed of water surface waves on the density of water, their wavelength, and surface tension is:
\[
v \propto \sqrt{\frac{\gamma}{\rho \lambda}}
\]
