Question

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

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

Answer 1

The correct option is B $1:2$

Vibrating frequency is given by
$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}=\frac{1}{2L}\sqrt{\frac{T}{\rho A}}$

Since no. of antinodes $n$ corresponds to the $n^{th}$ harmonic,
Frequency of vibration of wire $\text{A}$,
$f_A=\frac{p}{2L}\sqrt{\frac{T}{\rho \pi r^2}}$
and

Frequency of vibration of wire $\text{B}$,
$f_B=\frac{q}{2L}\sqrt{\frac{T}{\rho \pi (2r{)}^2}}\Rightarrow f_B=\frac{q}{4L}\sqrt{\frac{T}{\rho \pi r^2}}$

Since, $p^{th}$ harmonic of wire $\text{A}$ coincides with $q^{th}$ harmonic of wire $\text{B}$, we can say that, $f_A=f_B\Rightarrow \frac{p}{q}=\frac{1}{2}$