Here, L = 20 cm = 0.2 m, $\upsilon_n = 430$ Hz, $v = 340\,\text{m}\,\text{s}^{-1}$ The frequency of nth normal mode of vibration of closed pipe is $\upsilon_n = (2n-1)\frac{v}{4L}$ $\therefore 430 = (2n-1)\frac{340}{4 \times 0.2}$ or $2n-1 = \frac{430 \times 4 \times 0.2}{340} = 1.02$ $2n = 2.02$ or $n = 1.01$ Hence it will be the lst normal mode of vibration. In a pipe, open at both ends, we have $\upsilon_n = n \times \frac{v}{2L} \Rightarrow \frac{n \times 340}{2 \times 0.2} = 430$ $\therefore n = \frac{430 \times 2 \times 0.2}{340} = 0.5$As n has to be an integer, therefore, open organ pipe cannot be in resonance with the source.