For a fixed temperature, when the pressure varies, correspondingly density also varies such that the ratio \((\frac{p}{\rho})\) becomes constant. This means that the speed of sound is independent of pressure for a fixed temperature.

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 transverse waves, the distance between two neighbouring crests or troughs is known as the wavelength. For longitudinal waves, the distance between two neighbouring compressions or rarefactions is known as the wavelength. The SI unit of wavelength is meter.

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

(a) Bulk modulus
\(B =V\left|\frac{\Delta P}{\Delta V}\right|=\frac{100 \times 10^{3}}{0.005 \times 10^{-2}}= \)
\(=\frac{100 \times 10^{3}}{5 \times 10^{-5}}=2000 \mathrm{MPa}, \text { where }\)
MPa is mega pascal
(b) Speed of sound in water is
\(v=\sqrt { \frac { B }{ \rho } } =\sqrt { \frac { { 200\times 10 }^{ 6 } }{ 1000 } } =1414 \ ms^{-1}\)

	First case	Second case
(a) Wavelength	λ = 6 m	λ = 2 m
(b) Frequency	f = 2 Hz	f = 6 Hz
(c) Velocity	v = 6 × 2 = 12 m s-1	v = 2 × 6 = 12 m s-1

This means that the speed of the wave along a string is a constant. Higher the frequency, shorter the wavelength and vice versa, and their product is velocity which remains the same.

Answer is (c)

A jerk is given on a stretched string at time t = 0 s. Assume that the wave pulse created during this disturbance moves along positive x direction with constant speed v as shown in Figure (a).

Represent the shape of the wave pulse, mathematically as y = y(x, 0) = j(x) at time t = 0s. Assume that the shape of the wave pulse remains the same during the propagation. After some time t, the pulse moving towards the right and any point on it can be represented by x' (read it as x prime) as shown in Figure (b). Then
y(x, t) =j(x') =j(x - vt)
Similarly, if the wave pulse moves towards left with constant speed v, then y =j(x + vt). Both waves y =j(x + v!) and y =j(x - vt) will satisfy the following one dimensional differential equation known as the wave equation
\({∂^2y\over ∂x^2}={1\over4 v^2}{∂^2y\over ∂t^2}\)
where the symbol p represent partial derivative (read \({∂y\over ∂x}\)  as partial y by partial x). Not all the solutions satisfying this differential equation can represent waves, because any physical acceptable wave must take finite values for all values of x and t. But if the function represents a wave then it must satisfy the differential equation. Since, in one dimension (one independent variable), the partial derivative with respect to x is the same as total derivative in coordinate x, we write So it can be written as
\({d^2y\over dx^2}={1\over v^2}{d^2y\over dt^2}\)

(i) For the propagation of the waves, the medium must possess both inertia and elasticity, which decide the velocity of the wave in that medium.
(ii) In a given medium, the velocity of a wave is a constant whereas the constituent particles in that medium move with different velocities at different positions. Velocity is maximum at their mean position and zero at extreme positions.
(iii) Waves undergo reflections, refraction, interference, diffraction and polarization.

Given
Function \(a \rightarrow y=x^{2}+2 \alpha t x\)
Function \(b \rightarrow y=(x+\mathrm{vt})^{2}\)
Formula
For the function to be a wave function
\(\frac{d y / d x}{d y / d t}\) must be a constant.
For function (a)
\(y =x^{2}+2 \alpha t x  \)
\(\frac{d y}{d x} =2 x+2 \alpha t  \)   .....(1)
\(\frac{d y}{d t} =0+2 \alpha x=2 \alpha x\)   .......(2)
Dividing equation (1) by (2) we get
\(\frac{d y / d x}{d y / d t}=\frac{2 x+2 \alpha t}{2 d x} \text { is not a constant }\)
∴ Function (a) is not describing wave
For function b :
\(\mathrm{y} =(x+\mathrm{vt})^{2}  \)
\(\therefore \frac{d y}{d x} =2(x+\mathrm{vt}) \times 1=2(x+\mathrm{vt}) \)   .....(3)
\(\frac{d y}{d t} =2(x+\mathrm{vt}) \times \mathrm{v} \)
\(=2 \mathrm{v}(x+\mathrm{vt})\)  ......(4)
Dividing equation (3) by equation (4) we get
\(\frac{d y / \mathrm{dx}}{d y / \mathrm{dt}}=\frac{2(x+v t)}{2 v(x+v t)}=\frac{1}{v}=\text { constant }\)
Hence function b satisfies the wave equation.

From equation (11.25), we have
\(v=\sqrt { \frac { \gamma P }{ \rho } } \)
But \(\rho =\frac { M }{ V } \)
Therefore,\(v=\sqrt { \frac { \gamma PV }{ M } } \)
Using equation (11.26)
\(v=\sqrt { \frac { \gamma RT }{ M } } \) 
Where, R is the universal gas constant and M is the molecular mass of the gas. The speed of sound in nitrogen gas at 17°C is
\({ v }_{ N }=\sqrt { \frac { \gamma R\left( 273K+17K \right) }{ { M }_{ N } } } \)
\(=\sqrt { \frac { \gamma R(290K) }{ { M }_{ N } } } \\ \)............(1)
Similarly, the speed of sound in oxygen gas at temperature t
\({ v }_{ 0 }=\sqrt { \frac { \gamma R\left( 273K+t \right) }{ { M }_{ 0 } } } \) ............ (2)
Given that the value of γ is same for both the gases, the two speeds must be equal. Hence, equating equation (1) and (2), we get
\({ v }_{ 0 }={ v }_{ n }\)
\(\sqrt { \frac { \gamma R\left( 273+t \right) }{ { M }_{ 0 } } } =\sqrt { \frac { \gamma R(290K) }{ { M }_{ N } } } \)
Squaring on both sides and cancelling γ R term and rearranging, we get 
\(\frac { { M }_{ 0 } }{ { M }_{ N } } =\frac { 273+t }{ 290 } \)   ...............(3)
Since the densities of oxygen and nitrogen is 16:14,
\(\frac { { \rho }_{ 0 } }{ { \rho }_{ N } } =\frac { 16 }{ 14 } \)    ...(4)
\(\frac { { \rho }_{ 0 } }{ { \rho }_{ N } } =\frac { \frac { { M }_{ 0 } }{ V } }{ \frac { { M }_{ N } }{ V } } =\frac { { M }_{ 0 } }{ { M }_{ N } } \Rightarrow \frac { { M }_{ 0 } }{ { M }_{ N } } =\frac { 16 }{ 15 } \)   ....(5)
Substituting equation (5) in equation (3), we get
\(\frac { 273+t }{ 290 } =\frac { 16 }{ 14 } \Rightarrow 3822+14t=4640\)
\(\Rightarrow \) t = 58.4 °C

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

Here,
y = 2.0 cos 2π(8t - 0.0060x + 0.27)
= 2.0 cos [2π (8t - 0.0060x) + 2π(0.27)]
Standard equation for a travelling wave is,
\(y=r\ cos\left[ {2\pi\over \lambda}(vt-x)+\phi\right]\)
Here
\(\phi={2\pi\over \lambda}x=2\pi\times0.006x\)
\({2\pi\over \lambda}=0.006\)
(a) When x = 300 em,
ψ= 2π\(\times\) 0.006\(\times\)300
=3.6π rad.
(b) When x = 0.75 m = 75 em,
ψ = 2π\(\times\)0.006\(\times\)75
= 0.9π rad
(c) When \(x={{\lambda}\over 4}\)
\(\phi={2\pi\over \lambda}\times{\lambda\over4}={\lambda \over 2}rad\)

Newton assumed that when sound propagates in air, the formation of compression and rarefaction takes place in a very slow manner so that the process is isothermal in nature. That is, the heat produced during compression (pressure increases, volume decreases), and heat lost during rarefaction (pressure decreases, volume increases) occur over a period of time such that the temperature of the medium remains constant. Therefore, by treating the air molecules to form an ideal gas, the changes in pressure and volume obey Boyle's law, Mathematically
PV = Constant ..... (1)
Differentiating equation (1), we get
PdV + VdP = 0
or P = \(-V\frac{dP}{dV}=B_T\) ............... (2)
where, BT is an isothermal bulk modulus of air.
Substituting equation (2) in equation v = \(\sqrt{\frac{B}{p}},\)
V_r = \(\sqrt{\frac{B_r}{\rho}}=\sqrt{\frac{p}{\rho}}\) ............... (3)
Since P is the pressure of air whose value at NTP (Normal Temperature and Pressure) is 76 em of mercury, we have
P = (0.76\(\times\)13.6\(\times\)10³ \(\times\)9.8) N m^-2
p = 1.293 kg m^-3. here p is density of air
Then the speed of sound in air at Normal Temperature and Pressure (NTP) is
V_r = \(\sqrt{\frac{0.76\times 13.6\times 10^3\times 9.8}{1.293}}\)
= 279.80 ms¹ \(\approx\) 280 ms^-1 (theoretical value)
But the speed of sound in air at 0°C is experimentally observed as 332 ms^-1 which is close upto 16% more than theoretical value (Percentage error is \(\frac{(332-280)}{332}\times 100\%=15.6\%\))
This error IS not small.
Laplace assumed that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson's law (not Boyle's law as Newton assumed), which is
\(PV_{\lambda}\) = constant ............... (4)
where, \(\lambda=\frac{C_p}{C_v}\) which is the ratio between specific heat at constant pressure and specific heat at constant volume.
Differentiating equation (4) on both the sides,
we get
\(V_\gamma dP+P(\gamma V_\gamma^{-1}dV)=0\)
or \(\gamma P=-V\frac{dP}{dV}B_A\) .................. (5)
where, BA is the adiabatic bulk modulus of air. Now, substituting equation (5) in equation \(v=\sqrt{\frac{B}{\rho}},\) the speed of sound in air is
\(V_A=\sqrt{\frac{B_A}{\rho}}=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\gamma v_f}\)
V_A = 331 ms^-1.