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:

\[a = - 1, b = \frac{1}{2} \text{ and } c = \left( - \frac{1}{2} \right)\] 
∴ Frequency, f = kL⁻¹ F^1/2m^−1/2

\[= \frac{k}{L} F^{1/2} m^\left( - \frac{1}{2} \right) = \frac{k}{L} \sqrt{\left( \frac{F}{m} \right)}\]
\[f = \frac{k}{L} \sqrt{\left( \frac{F}{m} \right)}\]