Question

Two stretched wires $A$ and $B$ of the same lengths vibrate independently. If the radius, density and tension of wire $A$ are respectively twice those of wire $B$, then the fundamental frequency of vibration of $A$ relative to that of $B$ is

• A. $1:1$
• B. $1:2$
• C. $1:4$
• D. $1:8$

Answer 1

The correct option is B $1:2$

$r_A=2r_B$

${\mu }_A=2{\mu }_B$

$T_A=2T_B$

Fundamental frequency

$F_A=\frac{1}{2L}\sqrt{\frac{T_A}{{\mu }_A}}=\frac{1}{2L}\sqrt{\frac{{2T}_B}{{2\mu }_B}}=\frac{1}{2L}\sqrt{\frac{T_B}{{\mu }_B}}$

$F_B=\frac{1}{2L}\sqrt{\frac{T_B}{{\mu }_B}}$

$\Rightarrow F_A=F_B$

$\therefore [\frac{F_A}{F_B}=\frac{1}{1}=1:1]$

Hence Option (A) is correct.