Question

A open organ pipe of 80cm long is sounded. If the velocity of sound is 340 m/s .what is the fundamental frequency of vibration of the air cloumn

Answer 1

Answer:

Given:

Open organ pipe = 80 cm long

Velocity of sound = 340 m/s

To find:

Fundamental frequency in the organ pipe

Concept:

A fundamental standing wave in an open organ pipe will form nodes at closed end and an anti-node at open end.

Distance between adjacent nodes and anti-node = λ/4

Diagram:

Please see attached photo to understand better.

Calculation:

Let length be denoted as L

$\sf{ \red{ \therefore \: L = \dfrac{ \lambda}{4} }}$

$\sf{\implies \: \lambda =4 L}$

$\sf{frequency = \dfrac{velocity}{wavelength}}$

$\sf{frequency = \dfrac{V}{4L}}$

$\sf{frequency = \dfrac{340}{4×\frac{80}{100}}}$

$\sf{\implies \: frequency = 106.25 \: hz}$

Answer 2

$\huge{\underline{\underline{\bf{Solution}}}}$

$\rule{200}{2}$

$\tt Given\begin{cases} \sf{Length \: of \: organ \: open \: pipe = 80 \: cm} \\ \sf{Velocity \: of \: sound = 340 \: ms^{-1}} \end{cases}$

$\rule{200}{2}$

$\Large{\underline{\underline{\bf{To \: Find :}}}}$

We have to find the frequency in open organ pipe.

$\rule{200}{2}$

$\Large{\underline{\underline{\bf{Explanation :}}}}$

Let length be x.

We know that,

$\Large{\star{\boxed{\rm{x = \frac{\lambda}{4}}}}}$

_______________[Put Values]

$\tt{→\lambda = 4x}$

$\rule{150}{2}$

Now,

$\Large{\star{\boxed{\rm{Frequency = \dfrac{Velocity}{Wavelength}}}}}$

________________[Put Values]

$\tt{→ Frequency = \frac{V}{4x}} \\ \\ \tt{→ Frequency = \dfrac{340 \times 100}{4 \times 80}} \\ \\ \tt{→ Frequency = \dfrac{34000}{320}} \\ \\ \tt{→Frequency = 106.25 \: hz }$

$\sf{\therefore \: Frequency = 106.25 \: hz}$