Answer
$323.74\;Hz$
Work Step by Step
As the transverse waves are set up on both wires by an external source, the frequency on both wire will be same: $f_1=f_2$
For Aluminum wire: $f_1=n_1\frac{v_1}{2L_1}$
For Steel wire: $f_2=n_2\frac{v_2}{2L_2}$
for both cases, tension in the string is same. $T_1=T_2=mg$
Therefore, $v_1=\sqrt {\frac{T_1}{\mu_1}}=\sqrt {\frac{mg}{M_1/L_1}}=\sqrt {\frac{mg}{\rho_1 AL_1/L_1}}=\sqrt {\frac{mg}{\rho_1 A}}$
and, $v_2=\sqrt {\frac{T_2}{\mu_2}}=\sqrt {\frac{mg}{M_2/L_2}}=\sqrt {\frac{mg}{\rho_2 AL_2/L_2}}=\sqrt {\frac{mg}{\rho_2 A}}$
Substituting the above values, we obtain
$f_1=n_1\frac{1}{2L_1}\sqrt {\frac{mg}{\rho_1 A}}$ and $f_2=n_2\frac{1}{2L_2}\sqrt {\frac{mg}{\rho_2 A}}$
$\because\;\;f_1=f_2$
$n_1\frac{1}{2L_1}\sqrt {\frac{mg}{\rho_1 A}}=n_2\frac{1}{2L_2}\sqrt {\frac{mg}{\rho_2 A}}$
or, $\frac{n_1}{n_2}=\frac{L_1\sqrt {\rho_1}}{L_2\sqrt {\rho_2}}$
Substituting the given values
$\frac{n_1}{n_2}=\frac{L_1\sqrt {\rho_1}}{L_2\sqrt {\rho_2}}$
or, $\frac{n_1}{n_2}=\frac{60\times\sqrt {2.60}}{86.6\times\sqrt {7.80}}$
or, $\frac{n_1}{n_2}=0.40$
The lowest mode of frequency is generated in the standing wave having the joint as one of the nodes. To satisfy the ratio as well as the condition of lower frequency mode, the value of $n_1$ and $n_2$ should be $2$ and $5$ respectively.
Thus, substituting the known values, we obtain the lowest frequency:
$f_1=2\times\frac{1}{2\times60}\sqrt {\frac{10\times10^3\times 981}{2.60\times1.00\times10^{-2}}}\;Hz$
or, $f_1=323.74\;Hz$
Therefore, the lowest frequency is $323.74\;Hz$
