Question

A pipe of length 0.847 m is open at both ends. When the wind blows, the pipe resonates. If the air's temperature is 4.49°C, what is the frequency (in Hz) of the second overtone of this pipe?

How would the frequency you found in the previous question change if the temperature of the air was raised by several degrees?

Answer 1

Temperature in the pipe, $T= 4.49^\circ C$

Length of the pipe,

Then the frequency of second overtone of the pipe with both end open is given by,

$f= \frac{3v_{sound}}{2L}$

where velocity of sound in air depends on the temperature as,

$v_{s}= (331.3+0.606T(^\circ C))m/s$

Thus, the frequency of second overtone will become,

  $f= \frac{2(331.3+0.606T(^\circ C))}{2L}$..........................................(1)

Now, using the given values in above, we get

$f= \frac{3(331.3+0.606(4.49))}{2\times 0.847}= 591.53m$(ANS)

It can be seen from equation (1), that frequency of the wave is directly proportional to the temperature. Thus, when the temperature is increased by several degrees then frequency of the wave will also increase.