$A$ pipe $20 \; cm$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430 \; Hz$ source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is $340 \; m/s$).

(A) Length of the pipe,$l = 20 \; cm = 0.2 \; m$.
Speed of sound,$v = 340 \; m/s$.
Source frequency,$\nu = 430 \; Hz$.
For a pipe closed at one end,the frequency of the $n^{\text{th}}$ normal mode is given by $\nu_n = (2n - 1) \frac{v}{4l}$,where $n = 1, 2, 3, \dots$.
Substituting the values: $430 = (2n - 1) \frac{340}{4 \times 0.2} = (2n - 1) \frac{340}{0.8} = (2n - 1) \times 425$.
$2n - 1 = \frac{430}{425} \approx 1.01$.
$2n \approx 2.01 \implies n \approx 1$.
Thus,the first harmonic (fundamental mode) is resonantly excited.
For a pipe open at both ends,the frequency of the $n^{\text{th}}$ normal mode is given by $\nu_n = \frac{nv}{2l}$,where $n = 1, 2, 3, \dots$.
$n = \frac{2l \nu_n}{v} = \frac{2 \times 0.2 \times 430}{340} = \frac{172}{340} \approx 0.5$.
Since $n$ must be an integer,the source will not be in resonance with the pipe if both ends are open.