Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio $1: 2: 3: 4$.
Let, there are n number of loops in the string.
Length corresponding each loop is $\frac{\lambda}{2}$.
Now, we can write
$$\begin{aligned} &L=\frac{n \lambda}{2} \Rightarrow \lambda=\frac{2 L}{n}\quad\text { [for } n \text { loops] } \end{aligned}$$
$$\begin{array}{ll} \Rightarrow & \frac{v}{v}=\frac{2 L}{n} \Rightarrow[\because v=v \lambda] \\ \Rightarrow & v=\frac{n}{2 L} v=\frac{n}{2 L} \sqrt{\frac{T}{\mu}} \quad [\because \text { velocity of transverse waves }=\sqrt{T / \mu}] \end{array}$$
$$\begin{aligned} &\Rightarrow v \propto n\quad [\because \text { length and speed are constants] } \end{aligned}$$
$$\begin{aligned} &\text { So, }\\ &\begin{aligned} v_1: v_2: v_3: v_4 & =n_1: n_2: n_3: n_4 \\ & =1: 2: 3: 4 \end{aligned} \end{aligned}$$
The earth has a radius of 6400 km . The inner core of 1000 km radius is solid. Outside it, there is a region from 1000 km to a radius of 3500 km which is in molten state. Then again from 3500 km to 6400 km the earth is solid. Only longitudinal $(P)$ waves can travel inside a liquid. Assume that the $P$ wave has a speed of $8 \mathrm{~km} \mathrm{~s}^{-1}$ in solid parts and of $5 \mathrm{~km} \mathrm{~s}^{-1}$ in liquid parts of the earth. An earthquake occurs at some place close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth, if wave travels along diameter?
Speed of wave in solid = 8 km/s
$$\begin{aligned} &\text { Speed of wave in liquid }=5 \mathrm{~km} / \mathrm{s}\\ &\text { Required time }=\left[\frac{1000-0}{8}+\frac{3500-1000}{5}+\frac{6400-3500}{8}\right] \times 2 \quad[\because \text { diameter }=\text { radius } \times 2] \end{aligned}$$
$$=\left[\frac{1000}{8}+\frac{2500}{5}+\frac{2900}{8}\right] \times 2 \quad\left[\text { time }=\frac{\text { distance }}{\text { speed }}\right]$$
$$=[125+500+362.5] \times 2=1975$$
As we are considering at diametrically opposite point, hence there is a multiplication of 2.
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
$$\begin{aligned} & =\left(\frac{d y}{d t}\right)_{\max }=5(100 \pi) \times 1 \\ & =500 \pi \mathrm{~m} / \mathrm{s} \end{aligned}$$