Question

A wire of length $2L$, is made by joining two wires $A$ and $B$ of 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

Answer: (D)

Let mass per unit length of wires are ${\mu }_1$ and ${\mu }_2$ respectively. $\because$ Materials are same, so density $\rho$ is same.
${\mu }_1=\frac{\rho \pi r^{2}L}{L}=\mu$ and ${\mu }_2=\frac{\rho 4\pi r^{2}L}{L}=4\mu$
Tension in both are same $=T$, let speed of wave in wires are $V_1$ and $V_2$
$V_1=\sqrt{\frac{T}{\mu }}=V;V_2=\sqrt{\frac{T}{4\mu }}=\frac{v}{2}$
So fundamental frequencies in both wires are
$f_{01}=\frac{V_1}{2L}=\frac{V}{2L}\&f_{02}=\frac{V_2}{2L}=\frac{V}{4L}$
Frequency at which both resonate is L.C.M of
both frequencies i.e. $\frac{V}{2L}$.
Hence no. of loops in wires are 1 and 2 respectively.