Answer
The second lowest frequency at which $R_1$ and $R_2$ are in phase at D is $~~118~Hz$
Work Step by Step
We can find the length of the path traveled by $R_1$:
$\sqrt{d_2^2+d_3^2} = \sqrt{(20.0~m)^2+(12.5~m)^2} = 23.585~m$
We can find the length of the path traveled by $R_2$:
$\sqrt{(d_2+2d_1)^2+d_3^2} = \sqrt{[20.0~m+(2)(2.50~m)]^2+(12.5~m)^2} = 27.951~m$
We can find the path difference:
$\Delta L = 27.951~m-23.585~m = 4.366~m$
For the waves to be in phase at D, the path length difference could be $1.50~\lambda$
We can find $\lambda$:
$1.500~\lambda = 4.366~m$
$\lambda = \frac{4.366~m}{1.50}$
$\lambda = 2.91~m$
We can find the frequency:
$f = \frac{v}{\lambda}$
$f = \frac{343~m/s}{2.91~m}$
$f = 118~Hz$
The second lowest frequency at which $R_1$ and $R_2$ are in phase at D is $~~118~Hz$
