#### Answer

(a) $\lambda_1 = 2.42~m$
$\lambda_2 = 1.21~m$
$\lambda_3 = 0.807~m$
(b) $\lambda_1 = 4.84~m$
$\lambda_3 = 1.61~m$
$\lambda_5 = 0.968~m$

#### Work Step by Step

(a) For a tube that is open at both ends, the wavelengths are $\lambda_m = \frac{2L}{m}$
We can find the three longest wavelengths:
$\lambda_1 = \frac{2L}{1} = \frac{(2)(1.21~m)}{1} = 2.42~m$
$\lambda_2 = \frac{2L}{2} = \frac{(2)(1.21~m)}{2} = 1.21~m$
$\lambda_3 = \frac{2L}{3} = \frac{(2)(1.21~m)}{3} = 0.807~m$
(b) For a tube that is closed at one end, the wavelengths are $\lambda_m = \frac{4L}{m}$, where $m = 1, 3, 5,...$
We can find the three longest wavelengths:
$\lambda_1 = \frac{4L}{1} = \frac{(4)(1.21~m)}{1} = 4.84~m$
$\lambda_3 = \frac{4L}{3} = \frac{(4)(1.21~m)}{3} = 1.61~m$
$\lambda_5 = \frac{4L}{5} = \frac{(4)(1.21~m)}{5} = 0.968~m$