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:
$\nu \propto \sqrt{T}.....\left(\mathrm{i}\right)$

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)

$\frac{∆\nu }{\nu }=\frac{1}{2}\frac{∆T}{T}$