If $c$ is rms speed of molecules in a gas and $v$ is the speed of sound waves in the gas, show that $c / v$ is constant and independent of temperature for all diatomic gases.
We know that rms speed of molecules of a gas
$$c=\sqrt{\frac{3 p}{\rho}}=\sqrt{\frac{3 R T}{M}}\quad \text{.... (i)}$$
where $M=$ molar mass of the gas.
Speed of sound wave in gas $v=\sqrt{\frac{\gamma p}{\rho}}=\sqrt{\frac{\gamma R T}{M}}\quad \text{... (ii)}$
On dividing Eq. (i) by Eq. (ii), we get
$$\frac{c}{v}=\sqrt{\frac{3 R T}{M} \times \frac{M}{\gamma R T}} \Rightarrow \frac{c}{v}=\sqrt{\frac{3}{\gamma}}$$
where $\gamma=$ adiabatic constant for diatomic gas
$$\gamma=\frac{7}{5}\quad \left[\text { since } \gamma=\frac{C_p}{C_V}\right]$$
Hence, $$\frac{c}{v}=\text { constant }$$
Given below are some functions of $x$ and $t$ to represent the displacement of an elastic wave.
(i) $y=5 \cos (4 x) \sin (20 t)$
(ii) $y=4 \sin (5 x-t / 2)+3 \cos (5 x-t / 2)$
(iii) $y=10 \cos [(252-250) \pi t] \cos [(252+250) \pi t]$
(iv) $y=100 \cos (100 \pi t+0.5 x)$
State which of these represent
(a) a travelling wave along- $x$-direction
(b) a stationary wave
(c) beats
(d) a travelling wave along- $x$-direction
Given reasons for your answers.
(a) The equation $y=100 \cos (100 \pi t+0.5 x)$ is representing a travelling wave along $x$-direction.
(b) The equation $y=5 \cos (4 x) \sin (20 t)$ represents a stationary wave, because it contains sin, cos terms i.e., combination of two progressive waves
(c) As the equation $y=10 \cos [(252-250) \pi t] \cdot \cos [(252+250) \pi t]$ involving sum and difference of two near by frequencies 252 and 250 have this equation represents beats formation.
(d) As the equation $y=4 \sin (5 x-t / 2)+3 \cos (5 x-t / 2)$ involves negative sign with $x$, have if represents a travelling wave along $x$-direction.
In the given progressive wave $y=5 \sin (100 \pi t-0.4 \pi x)$ where $y$ and $x$ are in metre, $t$ is in second. What is the
(a) amplitude?
(b) wavelength?
(c) frequency?
(d) wave velocity?
(e) particle velocity amplitude?
Standard equation of a progressive wave is given by
$$y=\operatorname{asin}(\omega t-k x+\phi)$$
This is travelling along positive $x$-direction.
Given equation is $\quad y=5 \sin (100 \pi t-0.4 \pi x)$
Comparing with the standard equation
(a) Amplitude $=5 \mathrm{~m}$
(b) $k=\frac{2 \pi}{\lambda}=0.4 \pi$
$\therefore \quad$ Wavelength $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.4 \pi}=\frac{20}{4}=5 \mathrm{~m}$
(c) $\omega=100 \pi$
$$\begin{array}{rlrl} \omega =2 \pi v=100 \pi \\ \therefore \quad \text { Frequency } v =\frac{100 \pi}{2 \pi}=50 \mathrm{~Hz} \end{array}$$
$$\begin{aligned} &\text { (d) Wave velocity } v=\frac{\omega}{k} \text {, where } k \text { is wave number and } k=\frac{2 \pi}{\lambda} \text {. }\\ &\begin{aligned} & =\frac{100 \pi}{0.4 \pi}=\frac{1000}{4} \\ & =250 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { (e) }\quad y & =5 \sin (100 \pi t-0.4 \pi x) \quad \text{... (i)}\\ \frac{d y}{d t} & =\text { particle velocity } \end{aligned} \end{aligned}$$
From Eq. (i),
$$\frac{d y}{d t}=5(100 \pi) \cos [100 \pi t-0.4 \pi x]$$
For particle velocity amplitude $\left(\frac{d y}{d t}\right)_{\max }$
Which will be for $\{\cos [100 \pi t-0.4 \pi x]\}_{\max }=1$
$\therefore$ Particle velocity amplitude
$$\begin{aligned} & =\left(\frac{d y}{d t}\right)_{\max }=5(100 \pi) \times 1 \\ & =500 \pi \mathrm{~m} / \mathrm{s} \end{aligned}$$
For the harmonic travelling wave $y=2 \cos 2 \pi(10 t-0.0080 x+3.5)$ where $x$ and $y$ are in cm and $t$ is in second. What is the phase difference between the oscillatory motion at two points separated by a distance of
(a) 4 m
(b) 0.5 m
(c) $\frac{\lambda}{2}$
(d) $\frac{3 \lambda}{4}$ (at a given instant of time)
(e) What is the phase difference between the oscillation of a particle located at $x=100 \mathrm{~cm}$, at $t=T \mathrm{sec}$ and $t=5$ ?
$$\begin{aligned} &\text { Given, wave functions are }\\ &\begin{aligned} y & =2 \cos 2 \pi(10 t-0.0080 x+3.5) \\ & =2 \cos (20 \pi t-0.016 \pi x+7 \pi) \end{aligned} \end{aligned}$$
Now, standard equation of a travelling wave can be written as
$$y=a \cos (\omega t-k x+\phi)$$
On comparing with above equation, we get
$$\begin{aligned} & a=2 \mathrm{~cm} \\ & \omega=20 \pi \mathrm{rad} / \mathrm{s} \\ & k=0.016 \pi \end{aligned}$$
Path difference = 4 cm
$$ \begin{aligned} &\text { (a) Phase difference } \Delta \phi=\frac{2 \pi}{\lambda} \times \text { Path difference }\\ &\begin{aligned} \therefore \quad \Delta \phi & =0.016 \pi \times 4 \times 100 \quad \left(\because \frac{2 \pi}{\lambda}=k\right)\\ & =6.4 ~\pi \mathrm{~rad} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { (b) } \Delta \phi=\frac{2 \pi}{\lambda} \times(0.5 \times 100) \quad [\because \text { Path difference }=0.5 \mathrm{~m}] \\ =0.016 ~\pi \times 0.5 \times 100 \\ =0.8 \pi ~\mathrm{~rad} \end{aligned}$$
(c) $\Delta \phi=\frac{2 \pi}{\lambda} \times\left(\frac{\lambda}{2}\right)=\pi \mathrm{rad}\quad$ [$$\because$$ Path difference = $$\lambda/2$$]
(d) $\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4}=\frac{3 \pi}{2} \mathrm{rad}$
(e) $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{20 \pi}=\frac{1}{10} \mathrm{~s}$
$$\begin{aligned} \therefore \text { At } \quad x & =100 \mathrm{~cm} \\ t & =T \\ \phi_1 & =20 \pi T-0.016 \pi(100)+7 \pi \\ & =20 \pi\left(\frac{1}{10}\right)-1.6 \pi+7 \pi=2 \pi-1.6 \pi+7 \pi \quad \text{.... (i)} \end{aligned}$$
$$\begin{aligned} &\text { Again, at } x=100 \mathrm{~cm}, t=5 \mathrm{~s}\\ &\begin{aligned} \phi_2 & =20 \pi(5)-0.016 \pi(100)+7 \pi \\ & =100 \pi-(0.016 \times 100) \pi+7 \pi \\ & =100 \pi-1.6 \pi+7 \pi \quad \text{... (ii)} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\therefore \text { From Eqs. (i) and (ii), we get }\\ &\begin{aligned} \Delta \phi=\text { phase difference } & =\phi_2-\phi_1 \\ & =(100 \pi-1.6 \pi+7 \pi)-(2 \pi-1.6 \pi+7 \pi) \\ & =100 \pi-2 \pi=98 \pi \mathrm{rad} \end{aligned} \end{aligned}$$