For a closed organ Pipe

\(\ell=\frac{\lambda}{4} \)
\(\therefore \lambda=4 \ell\)
For a open organ pipe \(L=\frac{\lambda}{2} \quad \therefore \lambda=2 L\)
Fundamental frequency of a closed pipe
\(f_{c} =250 \mathrm{~Hz}  \)
\(f_{0} =\frac{V}{\lambda}  \)
\(=\frac{V}{2 L}\)
Fundamental frequency of open organ pipe
\(f_{o} =2\left(\frac{V}{4 L}\right)  \)
\(=2 \times f_{c}  \)
\(=2 \times 250=500 \mathrm{~Hz}\)
∴ Frequency of open organ pipe =500 Hz.

For a flute which is an open pipe, we have
Second harmonics f₂ = 2 f₁ = 900 Hz
Third harmonics f₃ = 3 f₁ = 1350 Hz
Fourth harmonics f₄ = 4 f₁ = 1800 Hz
For a clarinet which is a closed pipe, we have
Second harmonics f₂ = 3f₁ = 1350 Hz
Third harmonics f₃ = 5 f₁ = 2250 Hz
Fourth harmonics f₄ = 7f₁ = 3150 Hz

I₁ is the intensity of sound detected at a distance 3.0 m and it is given as 10^-2 W m^-2. Let I₂ be the intensity of sound detected at a distance 6.0 m. Then,
r₁ = 3.0 m, r₂ = 6.0 m
and since, I\(\propto \frac { 1 }{ { r }^{ 2 } } \)
the power output does not depend on the observer and depends on the baby. Therefore
\(\frac { { I }_{ 1 } }{ { I }_{ 2 } } =\frac { { r }_{ 2 }^{ 2 } }{ { r }_{ 1 }^{ 2 } } \)
\({ I }_{ 2 }{ =I }_{ 1 }\frac { { r }_{ 2 }^{ 2 } }{ { r }_{ 1 }^{ 2 } } \)
I₂ = 0.25\(\times\)10^-2 W m^-2

When two or more waves. superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency.
Number of beats per second
n = |f₁ - f₂| per second

Given two successive length (resonance) to be L₁ = 20 cm and L₂ = 85 cm
The frequency is f = 256 Hz
v = f \(\lambda\) = 2f \(\Delta\)L = 2f (L₂ − L₁)
= 2 × 256 × (85 − 20) × 10 −2 m s⁻¹
v = 332.8 cm⁻¹