Question

The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m, Guess the expression for its frequency from dimensional analysis.

Answer 1

Expression of frequency is given as follows:

Frequency,

$f\propto \left[L^a\right]...\left(1\right)$

$f\propto \left[F^b\right]...\left(2\right)$

$f\propto \left[M^c\right]...\left(3\right)$

Combining eAq(1) (2) and (3) we can say:

$f=\left[K{L}^a{F}^b{M}^c\right]$

where M = Mass / unit length

L = Length

F = Tension (Force)

Dimension of $.f=\left[T^{-1}\right]$

Dimension of right side:

Dimension of force, $F={\left[ML{T}^{-2}\right]}^b=\left[M^b{L}^b{T}^{-2b}\right]$

Dimension of mass per unit length, $={\left[M{L}^{-1}\right]}^c=\left[M^c{L}^{-c}\right]$

So, $\left[T^{-1}\right]$ = $\left[L^a\right]$ $\left[M^b{L}^b{T}^{-2b}\right]$ $\left[M^c{L}^{-c}\right]$

$\left[M^0{L}^0{T}^{-1}\right]$ =$=M^{b+c}{L}^{a+b-c}{T}^{-2b}$

Equating the dimensions of both sides, we get

$\text{b +c = 0}$ b +c = 0 ......(i)

$\text{- c + a + b = 0}$ - c + a + b = 0 ......(ii)

$\text{- 2 b = - 1}$ - 2 b = - 1 ......(iii)

Solving the equations, we get,

\$a=-1,b=\frac{1}{2}$ and $c=\frac{-1}{2}$

$\therefore f=K{L}^{-1}{F}^{\frac{1}{2}}{M}^{\frac{-1}{2}}$

Hence expression of frequency will be as follows:

$f=\frac{K}{L}\sqrt{\frac{F}{M}}$