Velocity of Waves in Liquid: Relation with Wavelength, γ & ρ

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The velocity of waves in a liquid is influenced by the wavelength (λ), surface tension (γ), and density (ρ) of the liquid. An equation can be derived to relate these variables, noting that surface tension is measured in force per unit length (N.m^-1). By analyzing the fundamental units, it becomes clear that combining λ, γ, and ρ results in units that require adjustment to achieve the desired velocity unit of m.s^-1. To resolve this, one of the variables must be placed in the denominator to cancel out the mass unit. Ultimately, the relationship can be expressed with a square root adjustment to ensure the units align correctly for wave velocity.
ahmedb
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The velocity of waves in a liquid depends on the wavelength λ, surface tension γ, and density ρ, of the liquid, Derive an equation to relate v, with wavelength, surface tension and density, (note: surface tension, has a formula of force per unlit length and SI unit of Nm^-1)
 
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ahmedb said:
The velocity of waves in a liquid depends on the wavelength λ, surface tension γ, and density ρ, of the liquid, Derive an equation to relate v, with wavelength, surface tension and density, (note: surface tension, has a formula of force per unlit length and SI unit of Nm^-1)

reducing to fundamental units

λ - m
γ - N.m-1 = kg.m.s-2.m-1 = kg.s-2
ρ - kg.m-3

The final answer is velocity, so m.s-1

If you multiply all three together [ie λ x γ x ρ] we have m x kg.s-2 x kg.m-3 = kg2.s-2.m-2

Not quite what we wanted. We needed the kg to cancel out, so one of γ or ρ needed to be in the denominator.

The fact that all units were squared doesn't matter - we can throw in a √ at the end to fix that.

have a play round and see how you get on.
 
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