Question

Show that if the room temperature changes by a small amount from T to T + ∆T, the fundamental frequency of an organ pipe changes from v to v + ∆v, where \[\frac{∆ v}{v} = \frac{1}{2}\frac{∆ T}{T} .\]

Answer 1

Let f  be the frequency of an open pipe at a temperature T. When the fundamental frequency of an organ pipe changes from v to v + ∆v, the temperature changes from T to T + ∆T.

We know that : 

\[\nu \propto   \sqrt{T}       .  .  .  .  . \left( i \right)\]

According to the question,

\[\nu +  ∆ \nu \propto   \sqrt{∆ T + T}\]

 Applying this in equation (i), we get:

\[\frac{\nu + ∆ \nu}{\nu} = \sqrt{\frac{∆ T + T}{T}}\]

\[1 + \frac{∆ \nu}{\nu} =  \left( 1 + \frac{∆ T}{T} \right)^{1/2} \]

By expanding the right-hand side of the above equation using the binomial theorem, we get:

\[1 + \frac{∆ \nu}{\nu} = 1 + \frac{1}{2} \times \frac{∆ T}{T}\] (neglecting the higher terms)

\[\frac{∆ \nu}{\nu} = \frac{1}{2}\frac{∆ T}{T}\]