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

Frequency, f $\propto$ L^aF^bm^c
f = kL^aF^bm^c ...(1)
Dimension of [f] = [T⁻¹]

Dimension of the right side components:
[L] = [L]
[F] = [MLT⁻²]
[m] = [ML⁻¹]

Writing equation (1) in dimensional form, we get:
[T⁻¹] = [L]^a [MLT⁻²]^b [ML⁻¹]^c
[M⁰L⁰T⁻¹] = [M^{b + c} L^{a + b − c} T^−2b]

Equating the dimensions of both sides, we get:
b + c = 0 ....(i)
a + b −c = 0 ....(ii)
−2b = −1 ....(iii)
Solving equations (i), (ii) and (iii), we get:
$a=-1,b=\frac{1}{2}\mathrm{and}c=\left(-\frac{1}{2}\right)$
∴ Frequency, f = kL⁻¹ F^1/2m^−1/2 $=\frac{k}{L}{F}^{1/2}{m}^{\left(-\frac{1}{2}\right)}=\frac{k}{L}\sqrt{\left(\frac{F}{m}\right)}$
⇒ $f=\frac{k}{L}\sqrt{\left(\frac{F}{m}\right)}$