Question

Four wires of identical lengths, diameters and of the same material are stretched on a sonometer wire. If the ratio of their tensions is 1 : 4 : 9 : 16, then ratio of their fundamental frequencies is

• A. 16 : 9 : 4 : 1
• B. 4 : 3 : 2 : 1
• C. 1 : 4 : 9 : 16
• D. 1 : 2 : 3 : 4

Answer 1

The correct option is D 1 : 2 : 3 : 4

Fundamental frequency is given by,
$f_0=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
Since, length ($l$), diameter (hence, cross sectional area) and material (hence, density) is same, therefore, mass per unit length ($\mu$) will also be same. Hence,
$f_0\propto \sqrt{T}$
Therefore,
$f_{0_1}:f_{0_2}:f_{0_3}:f_{0_4}=\sqrt{T_1}:\sqrt{T_2}:\sqrt{T_3}:\sqrt{T_4}$
$=1:2:3:4$