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.

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

Frequency, f = 900 MHz = 900\(\times\)10⁶ Hz
The speed of wave is c = 3\(\times\)10⁸ m s⁻¹
\(\lambda =\frac { v }{ f } =\frac { 3\times { 10 }^{ 8 } }{ { 900\times 10 }^{ 6 } } =0.33m\)

This implies, when increasing the value of a, the line shifts towards right side. For a = vt, y = x − vt satisfies the differential equation. Though this function satisfies the differential equation, it is not finite for all values of x and t. Hence, it does not represent a wave.

\({ \lambda }_{ 1 }=\frac { v }{ { f }_{ 1 } } =\frac { 340 }{ 20 } =17m\)
\({ \lambda }_{ 2 }=\frac { v }{ { f }_{ 2 } } =\frac { 340 }{ 20\times { 10 }^{ 3 } } =0.017m\)
Therefore, the audible wavelength region is from 0.017 m to 17 m when the velocity of sound in that region is 340 m s ^-1.

When a jerk is given to a stretched string which is tied at one end, a wave pulse is produced and the pulse travels along the string. Suppose two persons holding the stretched string on either side give a jerk simultaneously, then these two wave pulses move towards each other, meet at some point and move away from each other with their original identity. Their behaviour is very different only at the crossing/meeting points; this behaviour depends on whether the two pulses have the same or different shape as shown in Figure.

When the pulses have the same shape, at the crossing, the total displacement is the algebraic sum of their individual displacements and hence its net amplitude is higher than the amplitudes of the individual pulses. Whereas, if the two pulses have same amplitude but shapes are 1800 out of phase at the crossing point, the net amplitude vanishes at that point and the pulses will recover their identities after crossing. Only waves can possess such a peculiar property and It is called superposition of waves. This means that the principle of superposition explains the net behaviour of the waves when they overlap. Generalizing to any number of waves i.e, if two or more waves in a medium move simultaneously, when they overlap, their total displacement is the vector sum of the individual displacements.
To express mathematically, consider two functions which characterize the displacement of the waves, for example,
Y₁ = A₁ sin(kx - \(\omega t\))
and
Y₂ = A₂ cos(kx - \(\omega t\))
Since, both Y₁ and Y₂ satisfy the wave equation (solutions of wave equation) then their algebraic sum
Y = Y₁ + Y₂
also satisfies the wave equation. This means, the displacements are additive. Suppose we multiply Y₁ and y₂ with some constant then their amplitude is scaled by that constant Further, if C₁ and C₂ are used to multiply the displacernents y₁ andY₂ respectively, then, their net displacement Y is
Y = C₁Y₁ + C₂Y₂
This can be generalized to any number of waves. In the case of n such waves in more than one dimension the displacements are written using vector notation.
Here, the net displacement \(\vec y\) is
\(\vec { y } =\overset { n }{ \underset { i=1 }{ \Sigma } } { C }_{ i }\vec { { y }_{ i } } \)
The principle of superposition can explain the following:
(a) Space (or spatial) Interference (also known as Interference)
(b) Time (or Temporal) Interference (also known as Beats)
(c) Concept of stationary waves
Waves that obey principle of superposition are called linear waves (amplitude is much smaller than their wavelengths). In general, if the amplitude of the wave is not small then they are called non-linear waves.

The velocity of the wave
\(v=\sqrt { \frac { T }{ \mu } } \)
The length of the string, L = 80 cm=0.8 m
The mass of the string, m = 0.32 g =0.32 × 10^-3kg
Therefore, the linear mass density,
\(\mu =\frac { 0.32\times { 10 }^{ -3 } }{ 0.8 } =0.4\times { 10 }^{ -3 }{ kg\quad m }^{ -1 }\)
The tension in the string, T = 80 N
\(v=\sqrt { \frac { 80 }{ 0.4\times { 10 }^{ -3 } } } =447.2{ ms }^{ -1 }\)
The wavelength corresponding to the fundamental frequency f₁ is λ₁ = 2L = 2 × 0.8 = 1.6 m
The fundamental frequency f₁ corresponding to the wavelength λ₁
\({ f }_{ 1 }=\frac { v }{ { \lambda }_{ 1 } } =\frac { 447.2 }{ 1.6 } =279.5Hz\)
Similarly, the frequency corresponding to the second harmonics, third harmonics and fourth harmonics are
f₂ = 2f₁ = 559 Hz
f₃ = 3f₁ = 838.5 Hz
f₄ = 4f₁ = 1118 Hz

First harmonic of a closed organ pipe
\(\mathrm{L}_{\mathrm{c}}=\frac{\lambda}{4}\)
Third harmonic of an open organ pipe
\(\mathrm{L}_{\mathrm{o}} =\frac{3 \lambda}{2} \)
\(\frac{L_{c}}{L_{o}} =\frac{\lambda}{4} \times \frac{2}{3 \lambda} \)
\(\frac{L_{c}}{L_{o}} =, \frac{1}{6} \)

Factual information
\(\text { At } \mathrm{t}=0 \text { amplitude } \mathrm{y}=\frac{1}{1+x^{2}}\)
\(\text { At } \mathrm{t}=2 \text { amplitude } \mathrm{y}=\frac{1}{1+(x-2)^{2}}\)
\(\therefore v=\frac{\Delta y}{\Delta t} =\frac{2}{2} =1.0 \mathrm{~ms}^{-1} \)

No. of beats = 10
Frequency of pitch in the first wire
\(f_{1}=360 \mathrm{~Hz}\)
\(\therefore \text { Frequency of pitch in the second wire}\)
\(f_{2} =360-10=350 \mathrm{~Hz} \)
\(\text { Ratio of length }=x: y\)
\(l \propto f\)
\(\therefore \mathrm{x}: \mathrm{y} =\mathrm{f}_{1}: \mathrm{f}_{2} \)
\(=360: 350 =36: 35 \)

Frequency = 5000 Hz
Speed of sound in air = 332 m/s
Speed of sound in water = 1450 m/s
Wavelength of sound in air  \(=\frac{332}{5000}\)
\(=66.4 \times 10^{-3}\)
\(\text { Wavelength of sound water } \lambda_{\text {water }}\)
\(=\frac{1450}{5000} =290 \times 10^{-3} \mathrm{~m} \)
\(\therefore \frac{\lambda_{\text {water }}}{\lambda_{\text {air }}} =\frac{290 \times 10^{3}}{66.4 \times 10^{-3}} =4.367 \)

\(v_{\mathrm{A}}=500 \mathrm{~ms}^{-1} \quad \lambda_{A}=5 \mathrm{~m}\)
Frequency in medium B
\(\mathrm{f}_{\mathrm{B}}=\frac{v_{A}}{\lambda_{A}}=\frac{500}{5}=100 \mathrm{~Hz}\)
Wavelength in medium B
\(\lambda_{B}=\frac{v_{s}}{f_{s}}=\frac{600}{100}=6 \mathrm{~Hz}\)

In an Octane, let the frequencies of be f, (f + n), (f + 2n), ......f + (N - 7)n
∴ Sum of two frequencies = 2f
f + (N - 1) n = 2f 
(N - 1)n = 2f - f = f 
∴  f = n(N - 1)