Name: Closed Organ Pipe
Uploaded: 2019-07-19
Duration: 3 min 44 s
Description: An organ pipe is closed at one end and open at the other end. Bottle, whistle, etc., are examples of a closed organ pipe. Let's learn about its different mod...

Question

Explain graphically the formation of stationary wave as a result of reflection of a progressive wave from a rigid end.

Accepted Answer

Let us consider a closed organ pipe of length $l$ lying on the $x$ -axis, the closed-end at $x = 0$ and open end at $x = l$ . When air is flown at open end a longitudinal wave travels towards $x = 0$ which is reflected back from rigid and towards open end.

At the open end always antinode is formed and at closed end always node is formed.

The incident along $- v e$ direction of $x$ -axis is $y_{1} = asin \frac{2 π}{λ} (vt + x) . . . . . . . . . . . . . . . . . . (1)$

Where, $a =$ amplitude, $v =$ velocity and $λ =$ wavelength

Reflected wave suffers a phase change of $π$ at the rigid end and moves along positive $x -$ axis.

$∴ y_{2} = a s i n \frac{2 π}{λ} (v t - x + π) . . . . . . . . . . . . . . . . . . (2) = - a s i n \frac{2 π}{λ} (v t - x) . . . . . . . . . . . . . . . . . . (3)$

Superposition of $1$ and $3$ produce stationary waves

$y = y_{1} + y_{2} = 2 a \cos (\frac{2 πvt}{λ}) s i n \frac{2 π x}{λ}$

At closed end $x = 0$ , $∴ y = 0$ for all values of it.

Hence a node is formed at closed end.

At open end $x = l$

$∴ y = 2 a \cos (\frac{2 πvt}{λ}) s i n \frac{2 π l}{λ}$

Displacement $y$ will always be maximum, so $s i n \frac{2 π x}{λ} =$ maximum $= 1$

$\frac{2 π l}{λ} = (2 m - 1)$ where $m = 1, 2, 3, 4$

$λ = \frac{4 l}{2 m - 1} . . . . . . . . . . . . . . . . . . . . (4)$

Fisrt mode of vibration is also the fundamental node of vibration.

For, $m = 1, λ = 4 l ∴ l = \frac{λ}{4} . . . . . . . . . . . . . . . . . . . . . . . . . . (5)$

Hence in fig. $1 (a)$ antinode is formed at open end and node of closed and frequency and fundamental node is given by

$v = n λ = n 4 l ∴ n = \frac{v}{4 l}$

Second mode of vibration- Let $λ_{2}$ is the wavelength formed, then, $λ_{2} = \frac{4 l}{3}$ or $l = \frac{3 λ_{2}}{4}$

$n_{2}$ can be calculated as follows

$v = n_{2} λ v = n_{2} \frac{4 l}{3} ∴ n_{2} = \frac{3 v}{4 l} = 3 h$

Hence only odd harmonies are formed. First overtone and third harmonic shown in fig. $1$

Third mode of vibration, let $λ_{3}$ is the wavelength of the standing wave which mode forms $= 3$ .

Then, $λ_{3} = \frac{4 l}{3}$ from eq. $4$

$v = n_{3} λ_{3} \Rightarrow n_{3} = \frac{5 v}{4 l} = 5 h$

Hence only odd harmonics are formed is a pipe open at one end.

Explain graphically the formation of stationary wave as a result of reflection of a progressive wave from a rigid end.

Important Questions on Superposition of Waves-2 : Stationary (Standing) Waves : Vibration of Air Columns

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