Question

A wire of length $2L$, is made by joining two wires $A$ and $B$ of the same length, but different radii $r$ and $2r$ and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire $A$ is $p$ and that in $B$ is $q$ then the ratio $p:q$ is

• A. $4:9$
• B. $3:5$
• C. $1:4$
• D. $1:2$

Answer 1

The correct option is D $1:2$

Let mass per unit length of wires be ${\mu }_1$ and ${\mu }_2$ respectively.
Because the materials are the same, density $\rho$ is same.
$\therefore {\mu }_1=\frac{\rho \pi r^{2}L}{L}=\mu$ and ${\mu }_2=\frac{\rho 4\pi r^{2}L}{L}=4\mu$
Tension in both are the same $=T$
Let the speed of the wave in the wires be $V_1$ and $V_2$
$V_1=\sqrt{\frac{T}{\mu }}=V,V_2=\sqrt{\frac{T}{4\mu }}=\frac{V}{2}$
So fundamental frequencies in the wires are $f_{01}=\frac{V_1}{2L}=\frac{V}{2L}$ and $f_{02}=\frac{V_2}{2L}=\frac{V}{4L}$
Frequency at which both resonate is the L.C.M of both frequencies i.e. $\frac{V}{2L}$.
Hence, no. of loops in the wires are $1$ and $2$ respectively.
So, ratio of no. of antinodes is $1:2$