$A$ pipe,$30.0 \; cm$ long,is open at both ends. Which harmonic mode of the pipe resonates a $1.1 \; kHz$ source? Will resonance with the same source be observed if one end of the pipe is closed? Take the speed of sound in air as $330 \; m s^{-1}$.

(N/A) The first harmonic frequency for an open pipe is given by $v_1 = \frac{v}{2L}$.
Where $L$ is the length of the pipe and $v$ is the speed of sound.
The frequency of its $n^{th}$ harmonic is $v_n = \frac{nv}{2L}$ for $n = 1, 2, 3, \dots$
Given $L = 30.0 \; cm = 0.3 \; m$ and $v = 330 \; m s^{-1}$,the frequencies are:
$v_n = \frac{n \times 330}{2 \times 0.3} = \frac{330n}{0.6} = 550n \; Hz$.
For a source of $1.1 \; kHz = 1100 \; Hz$,we have $550n = 1100$,which gives $n = 2$.
Thus,the source resonates at the second harmonic.
If one end of the pipe is closed,the fundamental frequency is $v_1 = \frac{v}{4L} = \frac{330}{4 \times 0.3} = \frac{330}{1.2} = 275 \; Hz$.
In a pipe closed at one end,only odd harmonics are present ($v_n = n \times 275$ where $n = 1, 3, 5, \dots$).
The source frequency $1100 \; Hz$ would correspond to $n = \frac{1100}{275} = 4$.
Since $4$ is an even number,this mode is not possible for a pipe closed at one end. Therefore,no resonance will be observed.