Question

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

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

Answer 1

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

A sonometer wire is stretched between two fixed wedges.

$\Rightarrow$ Fundamental frequency of vibration,
$f=\frac{v}{2L}=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
$\Rightarrow f\propto \sqrt{T}$
Ratio of frequencies in wires considered is
$\therefore f_1:f_2:f_3:f_4=\sqrt{T_1}:\sqrt{T_2}:\sqrt{T_3}:\sqrt{T_4}$
Given that $T_1:T_2:T_3:T_4=1:4:9:16$
$\therefore f_1:f_2:f_3:f_4=\sqrt{T_1}:\sqrt{T_2}:\sqrt{T_3}:\sqrt{T_4}=1:2:3:4$