No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x and t, wave function must have a finite value. Out of the given functions for y none satisfies this condition. Therefore, none can represent a travelling wave.

Here, given v = 45 Hz, M = 35 x 10^{- 2} kg
\(\mu=\frac{\text { Mass }}{\text { Length }}=4.0 \times 10^{-2} \mathrm{kgm}^{-1}\)
\(l=\frac{M}{\mu}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}=\frac{7}{8} \mathrm{~m}\)
\(l=\frac{\lambda}{2}=\frac{7}{8} \Rightarrow \lambda=\frac{7}{4} \mathrm{~m}=1.75 \mathrm{~m}\)
(i) Speed \(v=v \times \lambda=45 \times 1.75=78.75 \mathrm{~m} / \mathrm{s}\)
(ii) As \(v=\sqrt{\frac{T}{\mu}} \Rightarrow T=v^{2} \times \mu\)
\(\Rightarrow T=(78.75)^{2} \times 4 \times 10^{-2} \Rightarrow T=248.06 \mathrm{~N}\) 

We know that, speed,  \(v\infty \sqrt { T } \)
By formula v = \(\frac { xRT }{ p } \)
Where T is in kelvin 
\(\frac { v_{ t } }{ v_{ 0 } } =\sqrt { \frac { 273+t }{ 273+0 } } =3\)
\(\\ \Rightarrow \frac { 273+t }{ 273 } =9\Rightarrow \ t=9\times 273-273=2184^{ 0 }C\)

(i) A short pip by a whistle
(a) will not have a fixed frequency.
(b) will not have fixed wavelength.
(c) will have the definite speed that will be equal to the speed of sound in air.
(ii) 0.05 Hz will be the frequency of repetition of the short pip.

Number of beats/s = (n₁ - n₂)
Hence, time interval between two successive beats = time interval between two successive maxima = \(\frac{1}{n_1−n_2}\)

In case of a string fixed at two ends, when the string vibrates in n loops
v_n= \(\frac{n}{2l}\sqrt{\frac{T}{μ}}\) ⇒v_n ∝ n
Hence, when the string vibrates in 1 loop, 2 loops, 3 loops, 4 loops, the frequencies are in the ratio 1:2:3:4.

The speed of sound in air is more at higher temperature, as v ∝ √T As we know, frequency v=v/λ is more, hence, v will be more and accordingly pitch will be more.

Water waves produced by a motorboat sailing in water are both longitudinal and transverse.

Let n₁ be the frequency of the closed pipe and n₂ of the open pipe and l₁, l₂ their corresponding lengths.
\(v=4 l_{1} n_{1}=2 l_{2} n_{2}\)
\(\text { or } \ n_{1}=\frac{v}{4 l_{1}} \text { and } n_{2}=\frac{v}{2 l_{2}}\)
Third overtone of the closed pipe (seventh harmonic) = 7n,
First overtone of the open pipe = 2n₂
\(\text { Given : } 7 n_{1}=2 n_{2} \text { or } \frac{7 v}{4 l_{1}}=\frac{2 v}{2 l_{2}} \text { or } \frac{l_{1}}{l_{2}}=\frac{7}{4}\)

The travelling harmonic wave is 
y ( x,t ) = 7.5 ( 0.0050 x + 12t + \(\frac{\pi}{4}\) )
At  x = 1 cm and  t = 1 s
y ( 1,1 ) = 7.5 sin ( 0.0050\(\times\) 1 x + 12\(\times\)1 + \(\frac{\pi}{4}\) ) 
= 7.5 sin (12.005 + \(\frac{\pi}{4}\) )  ......(i)
Now, \( \theta\) =  (12.005 + \(\frac{\pi}{4}\) ) rad
= \( \frac{180 }{\pi}\) ( 12.005 + \(\frac{\pi}{4}\) )  degree
= \( \frac{12.79 \times 180}{22/7}\)
= 732.35⁰
\(\therefore\) From Eq. (i), y(1,1) = 7.5 sin ( 732.55\(°\) )
= 7.5 sin (720 + 12.55\(°\))
= 7.5 sin 12.55\(°\)
=  7.5 \(\times\) 0.2173
= 1.63 cm
Velocity of oscillation 
v =\( \frac{d}{dt} \){ y ( 1,1 ) }
= \(\frac{d}{dt}\) [ 7.5 sin ( 0.005 x +12 t + \(\frac{\pi}{4}\) ) ]
= 7.5\(\times\) 12 cos [ 0.05 x + 12 t + \(\frac{\pi}{4}\) ]
At , x = 1cm, t = 1s
v = 7.5 \(\times\) 12cos  ( 0.05 x + 12  + \(\frac{\pi}{4}\) )
= 90 cos ( 732.55\(°\) )
= 90 cos( 720 + 12.55 )
v = 90 cos ( 12.55\(°\) )
=  90 \(\times\) 0.975
= 87.89 cm/s 
Comparing the given equations with the standard form 
\(\Rightarrow \) y ( x,t ) = a sin ( kx  = \(\omega\)t +\(\phi\)  )
We get, a = 7.5 cm , \(\omega\) = 12, 2\(\pi\) v = 12  or v = \(\frac{6}{\pi}\)
\(\therefore\)  \(\lambda = \frac{2\pi}{0.005}\)
= \(\frac{2 \times3.14}{0.005}\) 
= 12.56 m
Velocity of propogation ,
v  = v  \(\lambda\) 
= \( \frac{6}{\pi}\) \(\times\)12.56 m/s
= 24 m/s
We find that velocity at x = 1 cm, t = 1s is not equal to velocity of wave propagation . 
(ii) Now all points which are at a distance of \(\pm\) \( \lambda\) ,  \(\pm\)2 \( \lambda\),  \(\pm\) 3\( \lambda\) from x = 1 cm will have same transverse displacement and velocity.As \( \lambda\) = 12.56 m, therefore, all points at distances \(\pm \)12.6 m ,\(\pm \) 25.2 m , \(\pm \)37.8 m from x = 1cm will have same displacement and velocity, as x =1 cm point at t = 2s, 5s and 11s.

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

(i) Keen observer, curiosity and quest for knowledge.
(ii) As the bucket is filled with water, the length L of the air column above the water level goes on decreasing. This increase the frequency (v=v/4L) and hence, the pitch of sound produced.
(iii) (a) \(v=\frac { \sqrt { \Upsilon \rho } }{ \rho } \), velocity of sound is inversely proportional to the square root of density of gas.
(b) Velocity of sound in air is directly proportional to square root of absolute temperature \(\nu \delta \sqrt { T } \) 

495 Hz, 594 Hz, Yes

Given equation is y(x,t) = 3.0sin(36t + 0.018x + \(\pi /4\)) Comparing with standard equation
y(x,t) = \(asin(\omega t+kx+\phi )\)
\(\omega =36,\ k=\frac { 2\pi }{ \lambda } =0.018\)
\( \Rightarrow \lambda =\ least\ distance=\frac { 2\pi }{ \lambda } =\frac { 2\pi }{ 0.018 } cm\)
\( =\ 349.1\ cm\)

Given equation is y(x,t) = 3.0sin(36t + 0.018x + \(\pi /4\)) Comparing with standard equation
y(x,t) = \(asin(\omega t+kx+\phi )\)
By comparing amplitude, a=3cm
\(\Rightarrow 2\pi v=36\)
\(\Rightarrow v=\frac { 36 }{ 2\pi } =5.73Hz\)

v = 256 Hz, T= m x g, T = 90 x 9.8 = 882N
\(\mu =\frac { m }{ L } =8.0\times 10^{ -3 }kgm^{ -1 }\)
Amplitude, a = 5 cm = 0.05 cm
Velocity of the transverese wave
\(\Rightarrow v=\sqrt { \frac { T }{ \mu } } =\sqrt { \frac { 882 }{ 8\times 10^{ -3 } } } =3.32\times 10^{ 2 }m/s\)
\(\\ \omega =2\pi v=2\times 3.14\times 256\)
\(=\ 1.61\times 10^{ 3 }rad/s\)
\(\lambda =\frac { v }{ V } =\frac { 3.32\times 10^{ 2 } }{ 256 }\)
\( \\ k=\frac { 2\pi }{ \lambda } =\frac { 2\times 3.14\times 256 }{ 3.32\times 10^{ 2 } } =4.84m^{ -1 }\)
As wave propagation along positive X-axis
\(Y=\ asin(\omega t-kx)\)
\( =\ 0.05sin(1.61\times 10^{ 3 }t-4.84x)\)
Here x,y are in metre and t is in second.

From  Kinetic theory of gases, 
\(p=\frac { 1 }{ 3 } pc^{ 2 }\), where c is rms speed of molecules of gas.
\(\Rightarrow c=\sqrt { \frac { 3p }{ p } } \)
v = speed of sound in the gas = \(\sqrt { \frac { p }{ \Upsilon \rho } } \)
\(\Rightarrow from\ eqs.(i)\ and\ (ii)\)
\( \frac { c }{ v } =\sqrt { \frac { 3p }{ \rho } \times \frac { \rho }{ \Upsilon \rho } } =\sqrt { \frac { 3 }{ \Upsilon } } \)
Fer diatomic gases, 
\(\Upsilon =1.4=\ constant\)
\(\\ \Rightarrow \frac { c }{ v } =\sqrt { \frac { 3 }{ 1.4 } } =1.46 =\ constant\) 

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