Here, L = 20 cm = 0.2 m, υn=430 Hz, v=340ms–1
The frequency of nth normal mode of vibration of closed pipe is
υn=(2n−1)

v

4L

∴430=(2n−1)

340

4×0.2

or 2n−1=

430×4×0.2

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
υn=n×

v

2L

⇒

n×340

2×0.2

=430
∴n=

430×2×0.2

340

=0.5
As n has to be an integer, therefore, open organ pipe cannot be in resonance with the source.