Question

A steel wire of length $60cm$ and area of cross section ${10}^{-6}{m}^2$ is joined with an aluminium wire of length $45cm$ and area of cross section $3\times {10}^{-6}{m}^2$. The composite string is stretched by a tension of $80N$. Density of steel is $7800kg{m}^{-3}$ and that of aluminium is $2600kg{m}^{-3}$. The minimum frequency of tuning fork, which can produce standing wave in it with node at the joint is

• A. $357.3Hz$
• B. $375.3Hz$
• C. $337.5Hz$
• D. $325.5Hz$

Answer 1

Answer: (C)

$337.5Hz$

The lengths of the two strings are : $L^{\prime }=0.6m$ and $L^{\prime \prime }=0.45m$
The areas of cross section are : $A^{\prime }={10}^{-6}{m}^2$ and $A^{\prime \prime }=3\times {10}^{-6}{m}^2$
The densities are : ${\rho }^{\prime }=7800kg/m^3$ and ${\rho }^{\prime \prime }=2600kg/m^3$
Frequencies are $f^{\prime }$ and $f^{\prime \prime }$.
We know that frequency, $f=\frac{n}{2L}\left(\sqrt{\left(T/\mu \right)}\right)$
Consider $n=1,$
$f^{\prime }=84.4Hz$

$f^{\prime \prime }=112.5$
For any $n$,
$\frac{n^{\prime }}{n^{\prime \prime }}=\frac{L^{\prime }}{L^{\prime \prime }}\left(\sqrt{\left({\mu }^{\prime }/{\mu }^{\prime \prime }\right)}\right)$
=> $n^{\prime }/n^{\prime \prime }=4/3$
Therefore for a standing wave with a node at the junction, the frequency of the source should be $4f^{\prime }$ or $3f^{\prime \prime }$.
That is $f=337.5Hz$