Equation for a travelling harmonic wave is given as:
y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)
= 2.0 cos (20πt – 0.016πx + 0.70 π)
Where,
Propagation constant, k = 0.0160 π
Amplitude, a = 2 cm
Angular frequency, ω = 20 π rad/s
Phase difference is given by the relation:
\(\phi=k x=2 \frac{\pi}{\lambda}\)
(a) For x = 4 m = 400 cm
Φ = 0.016 π x 400 = 6.4 π rad
(b) For 0.5 m = 50 cm
Φ = 0.016 π x 50 = 0.8 π ra
(c) For\( x=\frac{\lambda}{2}\)
\(\phi=2 \frac{\phi}{\lambda} \times \frac{\lambda}{2}=\pi r a d\)
(d) For \(x=\frac{3 \lambda}{4}\)
\(\phi=2 \frac{\pi}{\lambda} \times 3 \frac{\lambda}{4}=1.5 \pi r a d\) 

The given equation is \(y(x,t)=0.06 sin \frac{2\pi}{3} x cos (120 \pi t)\)  --- (1)
(i) As the equation involves harmonic functions of x and t separately, it represents a stationary wave.
(ii) We know that when a wave pulse 
\(y_{1}=r sin \frac{2\pi}{\lambda}(vt-x)\)
travelling along + direction of x-axis is superimposed by the reflected wave
\(y_{2}= -r sin \frac{2\pi}{\lambda}(vt+x)\)
travelling in opposite direction, a stationary wave
\(y=y_{1}+y_{2} = -2r sin(\frac{2\pi}{\lambda})x cos \frac{2\pi}{\lambda}\)vt is formed.   --- (2)
Comparing eqns. (1) and (2), we find that
\(\frac{2\pi}{\lambda}=\frac{2\pi}{3} \Rightarrow \lambda =3m\)
Also, \(\frac{2\pi}{\lambda}v=120 \pi\) or \(v=60 \lambda = 60\times 3=180 ms^{-1}\)
Frequency, \(v=\frac{v}{\lambda}=\frac{180}{3}=60 Hz\)
Note that both the waves have same wavelength, same frequency and same speed.
(iii) Velocity of transverse waves is 
\(v=\sqrt{\frac{T}{m}}\) or \(v^{2}=\frac{T}{m}\)
T=mv², where \(m=\frac{3\times 10^{-2}}{1.5}=2 \times 10^{-2} kg/m\)
T=\((180)^{2}\times 2 \times 10^{-2}\)
= 648 N.

The transverse harmonic wave is
y (x, t) = 3.0 sin\((36t+0.018x+\frac{\pi}{4})\)
For x = 0,
y(0,t) = \(3 sin (36t+0+\frac{\pi}{4}) = 3 sin (36t+ \frac{\pi}{4})\)   --- (1)
Here \(\omega = \frac{2\pi}{T}=36 \Rightarrow T=\frac{2\pi}{36}\)
To plot a (y) versus (t) graph, different values of y corresponding to the values of t may be tabulated as under (by making use of eqn. (1)).

Using the values of t and y (as in the table), a graph is plotted as under The graph obtained is sinusoidal.

Similar graphs are obtained for x = 2 cm and x = 4 cm. The oscillatorymotion in the travelling wave only differs in respect of phase. Amplitude and frequency of oscillatory motion remains the same in all the cases.

(i) Node  It is a point where the amplitude of oscillation is zero, i.e. displacement is minimum. As pressure is inversely related with displacement, i.e. when displacement will be minimum, the pressure will be maximum.
Antinode At this point displacement is maximum, i.e. amplitude of oscillation will be maximum and hence, the pressure will be minimal as it is inversely related.
(ii) Bats emit ultrasonic waves of large frequencies. These waves will be reflected by the obstacles in their path. The reflected rays received by the bat will give an idea about the obstacle, i.e. distance, direction, size and nature.
(iii) As overtones produced and relatives strengths of notes are different in two notes of violin and sitar. Although frequencies are same, we will distinguish by their strengths.
(iv) The reason behind is that solids have both the elasticity of volume as well as shape, whereas gases have only the volume elasticity.
(v) As in the dispersive medium wavelengths are different, hence the velocities, therefore, the shape of the pulse gets distorted.

(i) Effect of pressure
 \(\text{v speed of sound in a gas}=\sqrt { \frac { \Upsilon p }{ \rho } } \)
\(\text{ p=pressure,} \rho = \text{density,} \frac { M }{ V } \Rightarrow V=\sqrt { \frac { \Upsilon pV }{ M } } \)
When T is constant, pV = constant \(\Rightarrow \) v = constant 
Hence, velocity of sound is independent of the change in pressure of the gas provided temperature remains constant
(ii) Formula for Velocity, \(v=\sqrt { \frac { \Upsilon p }{ \rho } } \)
According to standard gas equation,
\(pV=RT\Rightarrow \rho =\frac { RT }{ V } \Rightarrow v=\sqrt { \frac { \Upsilon \times RT }{ pV } } =\sqrt { \frac { \Upsilon RT }{ M } } \)
Where M = pV = molecule weight of the gas
\(\Rightarrow v\infty \sqrt { T } \)
Hence, v increases with temperature.
(iii) Due to pressure of water vapours in air density changes. Hence, velocity of sound changes with humidity.
Let \(\rho _{ m }\) = density of moist \( \rho _{ d }\) = density of dry air
 v_m = velocity of sound in moist air
v_d = velocity of sound in dry air
\( v_{ m }\ =\ \sqrt { \frac { \Upsilon p }{ \rho _{ m } } } .v_{ d }=\sqrt { \frac { \Upsilon p }{ \rho _{ d } } } \)
\(\\ \frac { v_{ m } }{ v_{ d } } =\sqrt { \frac { \rho _{ d } }{ \rho _{ m } } } as\rho _{ d }>p_{ m }\Rightarrow v_{ m }>v_{ d }\) 

The first harmonic frequency is given by
\(v_1=\frac{v}{\lambda_1}=\frac{v}{2 L} \) (open pipe) 
where L is the length of the pipe. The frequency of its nth harmonic is:
\(v_n=\frac{n v}{2 L} \text {, for } n=1,2,3, \ldots\)(open pipe) 
First few modes of an open pipe are shown in Fig.
For \(L=30.0 \mathrm{~cm}, V=330 \mathrm{~m} \mathrm{~s}^{-1}\),
\(v_{\mathrm{n}}=\frac{n 330\left(\mathrm{~m} \mathrm{~s}^{-1}\right)}{0.6(\mathrm{~m})}=550 \mathrm{n} \mathrm{s}^{-1}\)
Clearly, a source of frequency 1.1 kHz will resonate at v₂ , i.e. the second harmonic.


Now if one end of the pipe is closed (Fig.), it follows from Eq. that the fundamental frequency is
\(v_t=\frac{v}{\lambda_1}=\frac{v}{4 L}\)(pipe closed at one end) 
and only the odd numbered harmonics are present :
\(v_3=\frac{3 v}{4 L}, v_5=\frac{5 v}{4 L}\) and so on.
For L = 30 cm and v = 330 m s^–1, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed

On comparing this displacement equation with Eq.  y (x, t) = a sin (kx – ωt), we find
(a) the amplitude of the wave is 0.005 m = 5 mm.
(b) the angular wave number k and angular frequency ω are k = 80.0 m^–1 and ω = 3.0 s^–1
We, then, relate the wavelength λ to k through Eq. , λ = 2π/k
\(\lambda =\frac { 2\pi }{ k } =\frac { 2\pi }{ 80 m^{-1} }\)
= 7.85 cm
(c) Now, we relate T to ω by the relation T = 2π/ω
\( =\frac { 2\pi }{ 3.0 m^{-1} }\)= 2.09 s
and frequency, v = 1/T = 0.48 Hz
The displacement y at x = 30.0 cm and time t = 20 s is given by
y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20)
= (0.005 m) sin (–36 + 12π)
= (0.005 m) sin (1.699)
= (0.005 m) sin (970 ) ≃ 5 mm

(a) Transverse and longitudinal
(b) Longitudinal
(c) Transverse and longitudinal
(d) Longitudinal

We know that 1 mole of any gas occupies 22.4 litres at STP.
Therefore, density of air at STP is: ρ_o = (mass of one mole of air)/ (volume of one mole of air at STP)
\(\\ =\frac { 29.0\times { 10 }^{ -3 }kg }{ 22.4\times { 10 }^{ -3 } } \) = 1.29 kg m^–3
According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP,
v = \(\left[ \frac { 1.01\times 10^{ 5 }{ Nm }^{ -2 } }{ 1.29kgm^{ -3 } } \right] ^{ { 1 }/{ 2 } }\) = 280 m s^–1

Mass per unit length of the wire,
\(\mu  =\frac{5.0 \times 10^{-3} \mathrm{~kg}}{0.72 \mathrm{~m}} \)
\(=6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}\)
Tension, T=60 N
The speed of wave on the wire is given by
\(v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{60 \mathrm{~N}}{6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}}=93 \mathrm{~m} \mathrm{~s}^{-1}\)

Give, v = 1000 kHz =10⁶ Hz
v_a = 340 m/s, v_w = 1486 m/s
Wavelength of reflected sound, \({ \lambda }_{ a }=v\frac { { v }_{ a } }{ v } \)
\(=\frac { 340 }{ { 10 }^{ 6 } } =3.4\times { 10 }^{ -4 }m\)      
Wavelength of transmitted sound,
\({ \lambda }_{ w }=\frac { { v }_{ s } }{ v } =\frac { 1486 }{ { 10 }^{ 6 } } =1486\times { 10 }^{ -6 }\)     
\({ \lambda }_{ w }=1.486\times { 10 }^{ -3 }m\)

(i) Here, given v = 45 Hz, M = 3.5 x 10^-2kg     
\(\mu =\frac { Mass }{ Length } =4.0\times { 10 }^{ -2 }{ kgm }^{ -1 }\)
\( l=\frac { M }{ \mu } =\frac { 3.5\times { 10 }^{ -2 } }{ 4\times { 10 }^{ -2 } } =\frac { 7 }{ 8 } m l=\frac { \lambda }{ 2 } =\frac { 7 }{ 8 } \)
\( \Rightarrow \lambda =\frac { 7 }{ 4 } m=1.75m\)
\(\\ Speed,\ v=v\times \lambda \ =45\times 1.75\ =78.75m/s \)
(ii) \(As\ v=\sqrt { \frac { T }{ \mu } } \Rightarrow T={ v }^{ 2 }\times \mu \)
\(\Rightarrow T=\left( 78.75 \right) ^{ 2 }\times 4\times { 10 }^{ -2 }\)
\( \Rightarrow T=248.06\ N\) 

Increase in the tension of a string increases its frequency. If the original frequency of B (ν_B ) were greater than that of A (ν_A ), further increase in ν_B should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that ν_B < ν_A . Since ν_A – ν_B = 5 Hz, and ν_A = 427 Hz, we get ν_B = 422 Hz.

Conceptual question based on fundamentals of characteristics of travelling wave.
The converse is not true means if the function can be represented in the form  y = f( x \(\pm \) vt ), it does not necessarily express a travelling wave. As the essential condition for a travelling wave is that the vibrating particle must have finite displacement value for all x and t.
(i) For x = 0
If t \(\rightarrow \)0, then (x - vt)²\(\rightarrow \)0 which is finite, hence, it is a wave as it passes the two tests.
(ii) log \(\left( \frac { x+vt }{ x_{ 0 } } \right)\)
l\( \\ At\ x=0\ and\ t=0,\)
\( f(x,t)=log\left( \frac { 0+0 }{ x_{ 0 } } \right) \)
= log 0 \(\rightarrow\) not defined
Hence, it is not a wave. 
(iii) \(\frac { 1 }{ x+vt } \)
\( \\ For\ x=0,\ t=0,\ f(x)\rightarrow \infty\)
Though the function is of (x\(\pm\) vt) type still at x = 0, it is infinite, hence, it is not a wave.

Given, l = 12.0m, M = 2\(\sqrt { \frac { T }{ \mu } } \).10kg, T = ?, v = 343m/s
 \(\mu \) = mass per unit length
=  \(\frac { M }{ l } =\frac { 2.10 }{ 12 } =0.175 \ kgm^{ -1 }\)
Also we know that, v = \(\sqrt { \frac { T }{ \mu } } \)
\(\Rightarrow T=v^{ 2 }\mu =(343)^{ 2 }\times (0.175)\)
\(\Rightarrow T=2.06\times 10^{ 4 }N\)

(i) All the points except the nodes on the string have the same frequency and phase but not the same amplitude.
(ii) Given, \(Y=0.06\sin { \frac { 2\pi }{ 3 } } x\cos { \left( 120\pi \right) } \)
Putting x = 0.375 m
Amplitude, \(Y=0.06\sin { \frac { 2\pi }{ 3 } } \times \left( 0.375 \right) \)
\(=0.06\sin { \frac { \pi }{ 4 } } =\frac { 0.06 }{ \sqrt { 2 } } =0.042\ m\)

As, there is piston at one end, it behaves as a closed organ pipe. Hence, it will produce odd harmonics only.
Hence, resonant frequencies will be first and third harmonics.
In the fundamental mode,\(\frac { \lambda }{ 4 } =25.5\ cm\) 
\(\Rightarrow \lambda =4\times 25.5\ =102\ cm=1.02\ m\)
Speed of sound in air
\(v=V\lambda =340\times \left( 1.02 \right) =346.8m/s\)  

Frequency of string A, f_A​=324Hz
Frequency of string B=f_B​
Beat’s frequency, n=6Hz
Beat's Frequency is given as:
n=∣f_A​−f_B​∣
6=∣324−f_B​∣
f_B​=330Hz or 318Hz
Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:
f ∝ \(\sqrt{T​}\)
Hence, the beat frequency cannot be 330 Hz
∴ f_B=318Hz

Speed of sound in the tissue, v = 1.7 km/s =1.7 × 10³m/s
Operating frequency of the scanner, ν = 4.2
MHz = 4.2 × 10⁶Hz
The wavelength of sound in the tissue is given as:
λ=v/ν
\(​​​​​​​=\frac{1.7×10^{34}}{2×10^{26}}\)
=4.1 × 10−4m

\(\mathrm{M}=2.50 \mathrm{~kg} \)
\( \mathrm{~T}=200 \mathrm{~N} \)
\( \mathrm{l}=20.0 \mathrm{~m}\)
Mass per unit length, \(\mu=\mathrm{M} / 1=2.50 / 20=0.125 \mathrm{Kg} \mathrm{m}^{-1}\)
The velocity (v) of the transverse wave in the string is given by the relation:
\( \mathrm{v}=\sqrt{\mathrm{T} / \mu} \)
\( =\sqrt{200 / 0.125}=\sqrt{1600}=40 \mathrm{~m} / \mathrm{s}\)
\(\therefore\) Time taken by the disturbance to reach the other end,
\(\mathrm{t}=1 / \mathrm{v}=20 / 40=0.5 \mathrm{~s}\)

Here, L = 100 cm = 1m, v = 2.53 k Hz = \(2.53 \times 10^{3}\)Hz.
When the rod is clamped at the middle, then in the fundamental mode of vibration of the rod, a node is formed at the middle and antinode is formed at each end.
Therefore, as is clear from Fig.

\(L=\frac{\lambda}{4}+\frac{\lambda}{4}=\frac{\lambda}{2}\)
λ = 2L = 2m
As \(\upsilon \)- = vλ
∴ \(\upsilon = 2.53 \times 10^{3} \times 2\)
= \(5.06 \times 10^{3} ms^{-1}\) 

(a) It represents a stationary wave.
(b) It does not represent either a travelling wave or a stationary wave.
(c) It is a representation for the travelling wave.
(d) It is a superposition of two stationary wave.

Given equation is y(x,t) = 3.0sin(36t + 0.018x + \(\pi /4\)) Comparing with standard equation
y(x,t) = \(asin(\omega t+kx+\phi )\)
(i) The given equation represents a transverse harmonic wave travelling from right to left (i.e., along negative X-axis). It is not stationary wave
By comparing, we get \(\omega \) = 36rad/s, k = 0.018/cm
Speed of wave, v = \(\frac { \omega }{ k } =\frac { 36 }{ 0.018 } 2000cm/s\)
(ii) By comparing amplitude, a = 3cm
\(\Rightarrow 2\pi v=36\)
\(\Rightarrow v=\frac { 36 }{ 2\pi } =5.73Hz\)
(iii) Initial Phase, \(\phi =\frac { \pi }{ 4 } \)
(iv) \(\omega =36,\ k=\frac { 2\pi }{ \lambda } =0.018\)
\(\Rightarrow \lambda =\ least\ distance=\frac { 2\pi }{ \lambda } =\frac { 2\pi }{ 0.018 } cm\)
\(=\ 349.1\ cm\) 

Given L = 20 cm = 0.2 m, v_n = 430 Hz, v = 340 m/s
It will behave as closed organ pipe
\( { v }_{ n }=\left( 2n-1 \right) \frac { v }{ 4L } ,\ where\ n=1,2,3....\)
\( \Rightarrow 430=\left( 2n-1 \right) \frac { v }{ 4L } =\left( 2n-1 \right) \times \frac { 340 }{ 4\times 0.2 } \)
\(\Rightarrow \ \left( 2n-1 \right) =\frac { 430\times \left( 0.8 \right) }{ 340 } \Rightarrow 2n=\frac { \left( 430 \right) \left( 0.8 \right) }{ 340 } +1\)
\( \Rightarrow n=\frac { 43\times 4 }{ 340 } +\frac { 1 }{ 2 } =\frac { 2\times 172+340 }{ 340\times 2 } =\frac { 684 }{ 680 } =1.006\)
Hence, it will be the 1st normal mode or harmonic mode of vibration.
In a pipe open at both ends,\({ v }_{ n }=n\times \frac { v }{ 2l } =\frac { n\times 340 }{ 2\times 0.2 } =430\) 
\(\Rightarrow n=\frac { 430\times 2\times 0.2 }{ 340 } =\frac { 43\times 2\times 2 }{ 340 } =0.5\)
As n is not integer, hence, open organ pipe cannot be in resonance with the source.

Given, h = 300m, g = 9.8m/s^2 , v = 340ms^-1
t₁ = time taken by stone to strike the water surface 
 \(t_{ 1 }=\sqrt { \frac { 2h }{ g } } =\sqrt { \frac { 300 }{ 49 } } =7.82s\left( as\quad h=0+\frac { 1 }{ 2 } gt^{ 2 }_{ 1 } \right) \)
t₂ = time taken by the splash's sound to reach top of the tower
\(t_{ 2 }=\frac { h }{ v } =\frac { 300 }{ 340 } =0.882\quad \left[ v=\frac { h }{ t_{ 2 } } \right] \)
Total time, t = time to hear splash of sound 
= t₁ + t₂ =7.82 + 0.882
= 8.702