Question

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

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

Answer 1

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

Fundamental frequency on the wire, $\nu =\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

where, $\mu ,T$ and $L$ are the mass per unit length, tension and length of the wire respectively.

Now for identical lengths, diameter and materials, $\nu \propto \sqrt{T}$

Thus, ${\nu }_1:{\nu }_2:{\nu }_3:{\nu }_4=\sqrt{T_1}:\sqrt{T_2}:\sqrt{T_3}:\sqrt{T_4}$

${\nu }_1:{\nu }_2:{\nu }_3:{\nu }_4=\sqrt{1}:\sqrt{4}:\sqrt{9}:\sqrt{16}$

$⟹{\nu }_1:{\nu }_2:{\nu }_3:{\nu }_4=1:2:3:4$