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

A

0.75 cm

long closed organ pipe is vibrating in its fundamental mode. If the temperature of the room be

15 ° C

and the speed of sound in air at this temperature be

340 {ms}^{- 1}

, then determine the frequency of the note of the pipe. If the temperature rises to

25 ° C

, then what will be the frequency? Consider the effect of temperature on the length of the pipe negligible.

Accepted Answer

Given:

Length of closed organ pipe $(l) =$ $0.75 m$

Mode of vibration $=$ Fundamental

Temperature $= 15 ° C$

Speed of sound $(v)$ at $15 ° C = 340 m / s$

Frequency of sound in a closed organ pipe in fundamental mode:

$n = \frac{v}{4 l}$

Putting the given values, we get:

$n = \frac{340}{4 \times 0.75} = 113.3 Hz$

We know that speed of sound in air:

$(v) \propto T$ ......(i) where $(T)$ is temperature of air at absolute scale.

Initial temperature was $(273 + 15) K$ and the new temperature is $(273 + 25) K$ . Hence, using (i) for two given temperatures:

$\frac{v_{(15 + 273) K}}{v_{(25 + 273) K}} = \sqrt{\frac{288}{298}} \Rightarrow \frac{340}{v_{298 K}} = \sqrt{\frac{288}{298}} \Rightarrow v_{298 K} = 345.84 m / s$

New velocity of sound at increased temperature $= 345.84 m / s$

Hence, the fundamental frequency after the increase in temperature:

$∴ n = \frac{V}{4 l} = \frac{345.84}{4 \times 0.75} = \frac{345.84}{3.00} = 115.28 Hz$

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