The vibrations of four air columns are shown | vedclass

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(D) For figure $(a)$,$L = \frac{\lambda_1}{4} \Rightarrow \lambda_1 = 4L$.For figure $(b)$,$L = \frac{\lambda_2}{4} + \frac{\lambda_2}{2} + \frac{\lam

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$1$:$4$:$2$:$3$

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(D) For figure $(a)$,$L = \frac{\lambda_1}{4} \Rightarrow \lambda_1 = 4L$.For figure $(b)$,$L = \frac{\lambda_2}{4} + \frac{\lambda_2}{2} + \frac{\lambda_2}{4} = \lambda_2 \Rightarrow \lambda_2 = L$.For figure $(c)$,$L = \frac{\lambda_3}{4} + \frac{\lambda_3}{4} + \frac{\lambda_3}{2} = \lambda_3 \Rightarrow \lambda_3 = L$ (Note: Based on the diagram,it represents a pipe open at both ends with one node in the middle,so $L = \lambda_3/2 \Rightarrow \lambda_3 = 2L$).For figure $(d)$,$L = \frac{\lambda_4}{2} + \frac{\lambda_4}{4} = \frac{3\lambda_4}{4} \Rightarrow \lambda_4 = \frac{4L}{3}$.Since frequency $f \propto \frac{1}{\lambda}$,we have $f_1 : f_2 : f_3 : f_4 = \frac{1}{\lambda_1} : \frac{1}{\lambda_2} : \frac{1}{\lambda_3} : \frac{1}{\lambda_4}$.Substituting the values: $f_1 : f_2 : f_3 : f_4 = \frac{1}{4L} : \frac{1}{L} : \frac{1}{2L} : \frac{3}{4L}$.Multiplying by $4L$,we get $1 : 4 : 2 : 3$.

The vibrations of four air columns are shown below. The ratio of frequencies is

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