Question

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.

Answer 1

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

Then, $f\propto {\left[F\right]}^a{\left[l\right]}^b{\left[\mu \right]}^c$

or, $f=k{\left[F\right]}^a{\left[l\right]}^b{\left[\mu \right]}^c$ ...(i)

Here, k is a dimensionless constant.

Thus, $\left[f\right]={\left[F\right]}^a{\left[l\right]}^b{\left[\mu \right]}^c$

or, $\left[M^0{L}^0{T}^{-1}\right]={\left[ML{T}^{-2}\right]}^e{\left[L\right]}^b{\left[M{L}^{-1}\right]}^c$

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

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

Thus, $a+c=0$ ...(ii)

$a+b-c=0$ ...(iii)

$-2a=-1$ ...(iv)

Solving these three equations, we get

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

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

$f=k{\left(F\right)}^{1/2}{\left(l\right)}^{-1}{\left(\mu \right)}^{-1/2}orf=\frac{k}{l}\sqrt{\frac{F}{\mu }}$

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

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