Question

O1M.6 The phase speed l5 of a traveling transverse wave on a flexible string can plausibly depend only on two quantities, the tension force Fr on the string (which tells you how strongly the string is pulled back toward equilib- rium when it is disturbed) and the string's mass per unit length μ (which tells you about how long it takes the string to respond to a given restoring force). Assume that the speed depends on some product of powers of these quan- tities. Show that if these assumptions are true, dimensional ysis requires that (Q1.23) where C is some unknown unitless constant. (It turns out that C- 1, as problem Q1D.1 shows.)

Answer 1

The problem will be solved through the dimensional analysis.

The dimension of the physical quantities on the left-hand side must be equal to the dimensions on the right-hand side.

The dimensions of the given quantities

$\\ dim (|\vec v|)= LT^{-1}\\ dim(|\vec F_r|)= MLT^{-2}\\ dim(\mu)=ML^{-1}$

Let the empirical formula be

$\\ |\vec V|= (|\vec F_r|)^a\times (\mu)^b$.....................................................equation (A)

as c is dimensionless constant

Hence putting the dimensions we get

$\\ dim (|\vec v|)= (dim(|\vec F_r|)^a\times (dim(\mu))^b\\ \implies LT^{-1}=(MLT^{-2})^a\times (ML^{-1})^b\\ \implies LT^{-1}=M^{a+b}L^{a-b}T^{-2a-b}$

equation the powers of M, L and T on both the sides, we get

$\\a+b=0 ..................equation(1) \\a-b=1..................equation (2) \\ -2a-b=-1.............equation(3)$

adding equation 1 and 2 we get

$\\ 2a=1 \\ \implies a=1/2$

subtracting equation 2 from 1

$2b=-1 \\ \implies b=-1/2$

putting the values in equation (A) we get

$\\ |\vec v|= C(|\vec F_r|)^{1/2}\times (\mu)^{-1/2} \\ \\ |\vec v|= C\sqrt {\frac{|\vec F_r|}{\mu}}$

Hence we get the required relation.