The velocity v, of a wave in a stretched string, depends on the tension T, in the spring and the mass per unit length of the spring. Obtain an expression fo...

Answer Details

Dimensional derivation of the speed of a wave on a stretched string.

Assume the speed v depends on the tension T and the mass per unit length \(\mu\):

\[v=k\,T^{a}\mu^{b}\]

where k is a dimensionless constant. Writing the dimensions:

• \([v]=LT^{-1}\)
• \([T]=MLT^{-2}\) (force)
• \([\mu]=ML^{-1}\) (mass per unit length)

So \(LT^{-1}=(MLT^{-2})^{a}(ML^{-1})^{b}=M^{a+b}L^{a-b}T^{-2a}\).

Equating powers:

• M: \(a+b=0\)
• L: \(a-b=1\)
• T: \(-2a=-1\Rightarrow a=\tfrac{1}{2}\)

From these, \(b=-\tfrac{1}{2}\). Therefore

\[v=k\,T^{1/2}\mu^{-1/2}=k\sqrt{\dfrac{T}{\mu}}.\]

Thus the speed of the wave is proportional to the square root of the tension divided by the mass per unit length.