Answer to Question #237471 in Mechanics | Relativity for Ray

Question

The velocity v of a wave in a stretched string depends on tension T, in the string and the mass per unit length U of the string.Obtain an expression for V in terms of T and U, using the method of dimensions

Answer 1

Explanations & Calculations

• We can write the relationship between the quantities as follows

$\qquad\qquad \begin{aligned} \small v&\propto \small T^m.U^n \end{aligned}$

• Then comparing both sides for their dimensions, m and n can be found which helps writing a tentative relationship.

$\qquad\qquad \begin{aligned} \small [v]&=\small LT^{-1}\\ \small [T]&=\small MLT^{-2}\\ \small [U]&=\small \frac{[mass]}{[length]}=\frac{M}{L}=ML^{-1}\\ \\ \small L.H.S &: \small LT^{-1}\\ \small R.H.S &:\small (MLT^{-2})^m.(ML^{-1})^n\\ &:\small M^{(m+n)}.L^{(m-n)}.T^{(-2m)}\\ \\ \small index(L.H.S) &: \small index(R.H.S)\\ \small M \to 0&=\small m+n \implies m=-n\\ \small L \to 1&=\small m-n \\ &\implies \small m=\frac{1}{2},n=\frac{-1}{2}\\ \small T \to -2&=\small -2m\,(proves\,the\,results)\\ \\ \therefore\, v&\propto \small T^{\frac{1}{2}}.U^{\frac{-1}{2}}\\ &\propto\small \sqrt{\frac{T}{U}} \end{aligned}$

• Then a acceptable equation would be

$\qquad\qquad \begin{aligned} \small v&=\small k.\sqrt{\frac{T}{U}} \end{aligned}$

• Value of k to be found experimentally.