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You have an open air column of length 1.40 m in air at 24.2 °C. What is the frequency of the third harmonic?

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Answer:

f₃ = 371.35 Hz

Step-by-step explanation:

First we need to find the speed o sound at given 24.2°C. For that purpose, we use the formula:

v = v₀√(T/T₀)

where,

v = speed of sound at 24.2° C = ?

v₀ = velocity of sound at 0° C = 331 m/s

T = Given Temperature = 24.2° C + 273 = 297.2 k

T₀ = Reference Temperature = 0° C + 273 = 273 k

Therefore,

v = (331 m/s)√(297.2 k/273 k)

v = 345.36 m/s

Now, for the third harmonic in the open air column:

λ₃ = (2/3) L

where,

λ₃ = wavelength = ?

L = Length of open air column = 1.4 m

Therefore,

λ₃ = (2/3)(1.4 m)

λ₃ = 0.93 m

So, for the frequency of the third harmonic, we use the formula:

v = f₃ λ₃

f₃ = v/λ₃

where,

f₃ = frequency of third harmonic = ?

v = speed of sound at that temperature = 345.36 m/s

λ₃ = wavelength of third harmonic = 0.93 m

Therefore,

f₃ = (345.36 m/s)/0.93 m

f₃ = 371.35 Hz

User Andrei Zhytkevich
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