# Chapter 21 - Superposition - Exercises and Problems: 9

(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 $f_3$: $f_3 = 3~f_1 = (3)(25.0~Hz) = 75.0~Hz$

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