#### Answer

(a) $\lambda_1 = 4.8~m$
$\lambda_2 = 2.4~m$
$\lambda_3 = 1.6~m$
(b) $f_3 = 75.0~Hz$

#### Work Step by Step

(a) The wavelengths are $\lambda_n = \frac{2L}{n}$
We can find the three longest wavelengths:
$\lambda_1 = \frac{2L}{1} = \frac{(2)(2.4~m)}{1} = 4.8~m$
$\lambda_2 = \frac{2L}{2} = \frac{(2)(2.4~m)}{2} = 2.4~m$
$\lambda_3 = \frac{2L}{3} = \frac{(2)(2.4~m)}{3} = 1.6~m$
(b) It is given that $f_2 = 50.0~Hz$
We can find $f_1$:
$2~f_1 = f_2$
$f_1 = \frac{f_2}{2} = \frac{50.0~Hz}{2} = 25.0~Hz$
We can find $f_3$:
$f_3 = 3~f_1 = (3)(25.0~Hz) = 75.0~Hz$