Question

An open pipe of length 40 cm and diameter 5 cm vibrates. The wavelength of sound emitted by it in fundamental mode is


77 cm


81.5 cm


86 cm


78.5 cm

Answer 1

Given:

An open organ pipe of length 40 cm and diameter 5cm vibrates.

To find:

Wavelength of resonance in fundamental mode?

Calculation:

In this case, we need to APPLY END CORRECTION in case of resonance.

$\therefore L + \Delta L = \dfrac{ \lambda}{2}$

$\implies L + 0.6(d) = \dfrac{ \lambda}{2}$

Here 'd' stands for diameter of tube.

$\implies 40 + (0.6 \times 5) = \dfrac{ \lambda}{2}$

$\implies 40 + 3= \dfrac{ \lambda}{2}$

$\implies 43= \dfrac{ \lambda}{2}$

$\implies \lambda = 86 \: cm$

So, wavelength of sound in fundamental frequency is 86 cm.