If the length of a closed organ pipe is $1m $ and velocity of sound is $330 m/s$, then the frequency for the second note is
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(b)For closed pipe ${n_1} = \frac{v}{{4l}} = \frac{{330}}{4}$$Hz$
Second note = $3{n_1} = \frac{{3 \times 300}}{4}$$Hz$.
Second note = $3{n_1} = \frac{{3 \times 300}}{4}$$Hz$.
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