When the air inside a closed pipe is at a temperature of 20^ C , its fundamental frequency is 293 Hz . Determine its fundamental frequency if the temperature changes to 22^ C .

Step-by-step solution

The fundamental frequency of a closed pipe is given by f = v 4L . The speed of sound in a gas is proportional to the square root of its absolute temperature, v T . Therefore, the fundamental frequency is proportional to the square root of the absolute temperature, f T . Given T₁ = 20^ C = 293 K and f₁ = 293 Hz . The new temperature is T₂ = 22^ C = 295 K . Using the relation f₂ f₁ = T₂ T₁ : f₂ = 293 295 293 = 293 295 f₂ = (294 - 1)(294 + 1) = 294^2 - 1 f₂ 294 Hz