14. waves and sound — Physics STD 11 Science — Question

Length of pipe, $\mathrm{L}=17 \mathrm{cm}=17 \times 10^{-2} \mathrm{\,m}$

In a closed pipe (open at one end), the frequency of its $\mathrm{n}^{\text {th }}$ harmonic is

$\mathrm{v}_{\mathrm{n}}=\frac{\mathrm{nv}}{4 \mathrm{L}}$ where $\mathrm{n}=1,3,5,7,.........$

Let $\mathrm{n}^{\text {th }}$ harmonic of closed pipe resonates with $1.5 \,kHz$ source.

$\therefore v_{n}=1.5 \times 10^{3} \mathrm{\,Hz}$

$\therefore 1.5 \times 10^{3}=\frac{\mathrm{nv}}{4 \mathrm{L}}$

$\mathrm{n}=\frac{1.5 \times 10^{3} \times 4 \times 17 \times 10^{-2}}{340}$

$n=3$