Given :
l = 20 cm = 0.2 m
v = 340 ms-1
f = 430 Hz
For a pipe closed at one end :
f = \(\frac{(2n-1)v}{4l}\) n = 1, 2, 3 ……
⇒ 430 = (2n-1)\(\frac{340}{4 \times 0.2}\)
⇒ n = 1
⇒ Resonance occurs only for first/fundamental mode of vibration. For a pipe open at both ends,
f = nv/2l n = 1, 2, …..
⇒ 430 = \(\frac{n \times 340}{2 \times 0.2}\)
⇒ n = 0.51
Since n<1, resonance does not occur.