The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string and the mass per unit length $\mu$ of string. Derive the formula for frequency.

Suppose, that the frequency f depends on the tension raised to the power a, length raised to power b  and mass per unit length raised to power c. Then,

$$\begin{gathered} f\propto F^a{l}^b{\mu }^c \\ or, \\ f=k{F}^a{l}^b{\mu }^c \end{gathered}$$

Here, k is a dimensionless constant. Thus, 

$$\begin{gathered} \left[f\right]=\left[F{]}^a\right[\eta {]}^b[\mu {]}^c \\ or, \\ \left[{\mathrm{M}}^0{ \mathrm{L}}^0{ \mathrm{T}}^{-1}\right]={\left[{\mathrm{MLT}}^{-2}\right]}^a[ \mathrm{L}{]}^b{\left[{\mathrm{ML}}^{-1}\right]}^c \\ \mathrm{o}r, \\ \left[{\mathrm{M}}^0{ \mathrm{L}}^0{ \mathrm{T}}^{-1}\right]=\left[{\mathrm{M}}^{a+c}{ \mathrm{L}}^{a+b-c}{ \mathrm{T}}^{-2a}\right] \end{gathered}$$

For dimensional balance, the dimensions on both sides should be same.

Thus, $a+c=0\dots \text{ (ii) }$

$$\begin{gathered} a+b-c=0...\left(iii\right) \\ and \\ -2a=-1\dots \text{ (iv) } \end{gathered}$$

Solving these three equations, we get

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

Substituting these values in Eq. (i), we get

$$\begin{gathered} f=k\left(F{)}^{1/2}\right(l{)}^{-1}(\mu {)}^{-1/2} \\ or, \\ f=\frac{k}{l}\sqrt{\frac{F}{\mu }} \end{gathered}$$

Experimentally, the value of k is found to be $\frac{1}{2}$

Hence,

$f=\frac{1}{2l}\sqrt{\frac{F}{\mu }}$