The fundamental frequency of a closed pipe is 293 Hz when the air in it is a temperature of 20°C. What will be its fundamental frequency when the temperature changes to 22°C?

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Solution

Given:
The fundamental frequency of a closed pipe is 293 Hz. Let this be represented by f₁.
Temperature of the air in closed pipe T₁ = 20°C = 20 + 273 = 293 K
Let f₂ be the frequency in the closed pipe when the temperature of the air is T₂ .
∴ T₂= 22°C = 22 + 273 = 295 K

Relation between f and T:

$f \propto \sqrt{T}$
∴
$\frac{f_{1}}{f_{2}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}} \Rightarrow \frac{293}{f_{2}} = \frac{\sqrt{293}}{\sqrt{295}} \Rightarrow f_{2} = \frac{293 \times \sqrt{295}}{\sqrt{293}} = 294 Hz$

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