# Closed Organ Pipe and heating it

My text wrote that:

in a closed organ pipe as the temperature is increased then the number of beats may increase or decrease.

But I think that number of beats should increase only why would that decrease as neglecting end correction, $f =v / 4l$

where , $f$ would be fundamental frequency , $l$ is tube length, $v$ is velocity of sound.

Now as $v= \sqrt {\gamma R T / M}$

so increasing temperature would increase velocity and hence increase the frequency , and hence the number of beats should increase.

Why does the book says that it would increase or decrease ?

My text I think neglected the increase in length of pipe but would that make a difference ?

• Beat is not same as frequency Commented Apr 8, 2017 at 8:34
• its $f_2 - f_1$ which gives number of beats Commented Apr 8, 2017 at 8:52
• I know that but see what is root of 9 - root of 4 ....And now tell me what is root of 11 - root of 6 Commented Apr 8, 2017 at 8:59
• if say the temprature was made from 289 K to 324 K so velocity would increase by 18/17 and since $v/4l=f$ so it can be clearly seen , it also increases the root though is important here but if velocity increases by "x " times after taking root , frequency too increases "x " times Commented Apr 8, 2017 at 9:07
• Let someone else answer because as you see you just gave me temp beats may increase or decrease I don't know as your book suggests Commented Apr 8, 2017 at 9:11

## 1 Answer

When an organ pipe is heated the pipe will expand which will tend to decrease its frequency and the air inside it will heat upo thus increasing the speed of sound in the air which will tend to increase the frequency of the pipe.
The effect of heat the air on the frequency of the pipe is much greater than the increase in length of the pipe so the frequency of an organ pipe increases as temperature rises.

Assume that the frequency of a cold organ pipe is $400 \, \rm Hz$ and a beat of $4 \, \rm Hz$ is heard when a tuning fork is sounded.

Now the tuning fork could have one of two frequencies: $396 \, \rm Hz$ or $404 \, \rm Hz$.

If the tuning fork has a frequency of $396 \, \rm Hz$ and the temperature rises the beat frequency will increase because of the increase in frequency of the pipe.

If the tuning fork has a frequency of $404 \, \rm Hz$ and the temperature rises the beat frequency will decrease because of the increase in frequency of the pipe

A high enough temperature could result in no beat and an even higher temperature could eventually result in the beat frequency increasing.