A pipe of length 20 cm is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 425 Hz source? The speed of sound = 340 ms^-1

Let N be the frequency of the source and $n_p$ that of the $p$^th harmonic of a closed pipe. The source will resonantly excite that harmonic mode of the pipe for which:

 $N = v_p$     

Now for a closed pipe, we know that

 $n_p=\frac{pv}{4L}$   for any value of $p=1,3,5,.....$ 

Therefore·, for resonance,

$N=\frac{pv}{4L}$

$\Rightarrow p=\frac{4NL}{v}=\frac{4\times 425\times 0.2}{340}=1$