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

Answer 1

The correct option is C $1:2$


Let mass per unit length of the wires be ${\mu }_{A}and{\mu }_B$ respectively.

$\because$ For same material, density is also same,
So, ${\mu }_A=\frac{\rho \pi r^{2}L}{L}=\mu$
and ${\mu }_B=\frac{\rho 4\pi r^{2}L}{L}=4\mu$

Tension ($T$) in both connected wires is the same.

So, speed of the wave in the wires are

$V_A=\sqrt{\frac{T}{{\mu }_A}}=\sqrt{\frac{T}{\mu }}$

and $V_B=\sqrt{\frac{T}{{\mu }_B}}=\sqrt{\frac{T}{4\mu }}$

$(\because {\mu }_A=\mu$ and ${\mu }_B=4\mu )$

So, $n^{th}$ harmonic in such a wire system is given by

$f_n=\frac{nV}{2L}$

$\Rightarrow f_A=\frac{p{V}_A}{2L}=\frac{p}{2L}\sqrt{\frac{T}{\mu }}$
(for $p$ antinodes)

Similarly, $f_B=\frac{q{V}_B}{2L}=\frac{q}{2L}\sqrt{\frac{T}{4\mu }}=\frac{1}{2}\left(\frac{q}{2L}\sqrt{\frac{T}{\mu }}\right)$
(for $q$ antinodes)

As frequencies $f_A$ and $f_B$ will be equal,

So, $f_A=f_B\Rightarrow \frac{p}{2L}\sqrt{\frac{T}{\mu }}=\frac{q}{2L}\sqrt{\frac{T}{4\mu }}$

$\Rightarrow \frac{p}{q}=\frac{1}{2}$

or $p:q=1:2$