Answer to Question #152794 in Mechanics | Relativity for Abdul subhan

Question

Assuming that the wave speed on a stretch string depends on the linear mass density meu as v inversely proportional to f/mue use dimentianal analysis to show that a =1/2and b=1/2

Answer 1

Dimension of wave speed=[LT^-1]

Dimension of force=[MLT^-2]

Dimension of linear mass density=[ML^-1].

We have

$v=\frac{f^a}{\mu^b}$

Where v is called wave speed ,f is called force and $\mu$ is called linear mass density.Now putting dimension of each quantity in above equation

$[LT^{-1}]=\frac{[MLT^{-2}]^a}{[ML^{-1}]^b}$

$$[LT^{-1}] =[M^{a-b}L^{a+b}T^{-2a}]$$

On comparing both side we get

$-2a=-1$

$a=\frac{1}{2}$

And $a-b=0$

$a=b$

$b=\frac{1}{2}$ ($\because$ $a=\frac{1}{2}$ )