A closed organ pipe of length $1.2 \,\,m$ vibrates in its first overtone mode. The pressure variation is maximum at:
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in closed organ pipe in first overtone the pressure is maximum at $\frac{\lambda}{4}$ from the open end and total length will be equal to $\frac{\lambda \times 3}{4} .$ so,
$\frac{\lambda \times 3}{4}=1.2$
$\lambda=1.6 \mathrm{m}$
so pressure is maximum at $\frac{\lambda}{4}=0.4 \mathrm{m}$ from the open end
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