A pipe of length 85 cm is closed from one end. Find the number of...

fundamental frequency

$\text{L} = \dfrac{\lambda}{4}{v}_{1} = \dfrac{\text{v}}{4\text{L}}$



$\text{L} = \dfrac{\lambda}{2} + \dfrac{\lambda}{4} = \dfrac{3 \lambda}{4}$

${v}_{2} = \dfrac{3\text{v}}{4\text{L}}$

$∴ v = \text{odd}\left(\dfrac{\text{v}}{4\text{L}}\right)$

Now $\text{L} = 8 5 cm = \text{0.85 m}$

$\text{v} = 340 m/s$

$= \text{odd} \left(\dfrac{3 4 0}{4 \times \text{0.85}}\right)$

$= \text{odd} \times 100\text{Hz}$

Hence possible frequencies below 1250 Hz are

100 Hz, 300 Hz, 500 Hz...........1100 Hz

Hence 6