The fundamental frequency of a closed pipe is $293 H z$ when the air in it is at a temperature of $20^{o} C$ What will be its fundamental frequency when the temperature charges to $22^{o} C$ ?

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Solution

The fundamental frequency of the closed organ pipe at $20 ° C = f_{0}$ and at $22 ° C = f$
Since the frequency is proportional to the speed of sound V and the speed of sound is proportional to the square root of the temperature, hence

$\frac{f_{0}}{f} = \frac{\sqrt{T_{0}}}{\sqrt{T}}$

$\frac{293}{f} = \frac{\sqrt{293 K}}{\sqrt{295 K}}$

so, $f = 294 H z$

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The fundamental frequency of a closed pipe is 293 Hz when the air in it is at a temperature of 20oC What will be its fundamental frequency when the temperature charges to 22oC ?

The fundamental frequency of the closed organ pipe at 20°C=f0 and at 22°C=f Since the frequency is proportional to the speed of sound V and the speed of sound is proportional to the square root of the temperature, hence f0f=√T0√T293f=√293 K√295 Kso, f=294 Hz

The fundamental frequency of a closed pipe is $293 H z$ when the air in it is at a temperature of $20^{o} C$ What will be its fundamental frequency when the temperature charges to $22^{o} C$ ?