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

We know that for frequency f, T+Temperature f ∝√(T) So, f_1/f_2=√(T_1)/√(T_2) ⇒293/f_2=√(293)/√(295) ⇒ f_2 =293 ×√(295)/√(293)=294 Hz ...

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Standard XII

Physics

Dimensional Analysis

The fundament...

Question

The fundamental frequency of a closed pipe is 293 Hz when the air in it is at a temperature of $20^{\circ} C$ . What will be its fundamental frequency when the temperature changes to $22^{\circ} C$ ?

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Solution

We know that for frequency f,

T+Temperature

$f \propto \sqrt{T}$

So, $\frac{f_{1}}{f_{2}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}}$

$\Rightarrow \frac{293}{f_{2}} = \frac{\sqrt{293}}{\sqrt{295}}$

$\Rightarrow f_{2} = \frac{293 \times \sqrt{295}}{\sqrt{293}} = 294 H z$

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

We know that for frequency f, T+Temperature f∝√T So, f1f2=√T1√T2 ⇒293f2=√293√295 ⇒f2=293×√295√293=294Hz

The fundamental frequency of a closed pipe is 293 Hz when the air in it is at a temperature of $20^{\circ} C$ . What will be its fundamental frequency when the temperature changes to $22^{\circ} C$ ?

We know that for frequency f,

T+Temperature

$f \propto \sqrt{T}$

So, $\frac{f_{1}}{f_{2}} = \frac{\sqrt{T_{1}}}{\sqrt{T_{2}}}$

$\Rightarrow \frac{293}{f_{2}} = \frac{\sqrt{293}}{\sqrt{295}}$

$\Rightarrow f_{2} = \frac{293 \times \sqrt{295}}{\sqrt{293}} = 294 H z$