1   A string of mass $2.50 kg$ is under a tension of $200 N.$ The length of the stretched string is $20.0 m.$ If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Solution :

Mass of the string,$M=2.50 kg$$\\$ Tension in the string,$T=200 N$$\\$ Length of the string,$l=20.0 m$$\\$ Mass per unit length,$\mu =\dfrac{M}{l}\\ =\dfrac{2.50}{20}=0.125 kgm^{-1}$$\\$ The velocity (v) of the transverse wave in the string is given by the relation:$\\$ $v=\sqrt{\dfrac{T}{\mu}}\\ =\sqrt{\dfrac{200}{0.215}}\\ =\sqrt{1600}=40 m/s$$\\$ $\therefore $ Time taken by the disturbance to reach the other end,$\\$ $t=\dfrac{l}{v}=\dfrac{20}{40}=0.50 s$$\\$

2   A stone dropped from the top of a tower of height $300 m$ high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is $340 m s^{-1} ? ( g = 9. 8 m s^{-2} )$

Solution :

$3$ Height of the tower, $s = 300 m$$\\$ Initial velocity of the stone, $u = 0$$\\$ Acceleration, $a = g = 9.8 m / s^2$$\\$ Speed of sound in air$= 340 m/s$$\\$ The time $( t_1 )$ taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:$\\$ $ s=ut_1+\dfrac{1}{2}gt^2_1\\ 300=0+\dfrac{1}{2}*9.8*t^2_1\\ \therefore =\sqrt{\dfrac{300*2}{9.8}}=7.82 s$$\\$ Time taken by the sound to reach the tower,$t_2=\dfrac{300}{340}=0.88s$$\\$ Therefore, the time after which the splash is heard,$t=t_1+t_2$$\\$ $=7.82+0.88=8.7s$

3   A steel wire has a length of $12.0 m$ and a mass of $2.10 kg.$ What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at $20 ^o C = 343 m s^{-1}$ .

Solution :

Length of the steel wire,$l=12 m$$\\$ Mass of the steel wire,$m=2.10 kg $$\\$ Velocity of the transverse wave,$ v=343 m/s$$\\$ Mass per unit length,$\mu =\dfrac{m}{l}=\dfrac{2.10}{12}\\ =0.175 kgm^{-1}$$\\$ For tension $T,$ velocity of the transverse wave can be obtained using the relation:$v=\sqrt{\dfrac{T}{\mu}}\\ \therefore T=v^2\mu \\ =(343)^2*0.175=20588.57 \approx 2.06*10^4N$$\\$

4   Use the formula $v=\sqrt{\dfrac{\gamma P}{\rho}}$ to explain why the speed of sound in air$\\$ a. is independent of pressure,$\\$ b.increases with temperature,$\\$ c.increases with humadity.

Solution :

c. Let $v_m$ and $v_d$ be the speeds of sound in moist air and dry air respectively. Let $\rho_m$ and $\rho_d$ be the densities of moist air and dry air respectively.$\\$ Take the relation:$\\$ $v=\sqrt{\dfrac{\gamma P}{\rho}}$$\\$ Hence, the speed of sound in moist air is:$\\$ $v_m=\sqrt{\dfrac{\gamma P}{\rho_m}} ....(i)$$\\$ And the speed of sound in dry air is:$\\$ $v_d=\sqrt{\dfrac{\gamma P}{\rho_d}} ....(ii)$$\\$ On dividing equations (i) and (ii), we get:$\\$ $\dfrac{v_m}{v_d}=\sqrt{\dfrac{\gamma P}{\rho_m}*\dfrac{\rho_d}{\gamma P}}\\ =\sqrt{\dfrac{\rho_d}{\rho_m}} ....(ii)$$\\$ However, the presence of water vapour reduces the density of air, i.e.,$\\$ $\rho_d < \rho_m\\ \therefore v_m > v_d$$\\$ Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous medium, the speed of sound increases with humidity.

a. Take the relation:$v=\sqrt{\dfrac{\gamma P}{\rho}} ...(i)$$\\$ where,$\\$ Density,$\rho=\dfrac{\text{Mass}}{\text{Volume}}=\dfrac{M}{V}$$\\$ M=Molecular weight of gas $\\$ V=Volume of gas $\\$ Hence, equation (i) reduces to:$\\$ $v=\sqrt{\dfrac{\gamma PV}{M}} ....(ii)$$\\$ Now from the ideal gas equation for $n = 1:$$\\$ $PV = RT$$\\$ For constant $T, PV $= Constant $\\$ Since both $M$ and $\gamma$ are constants, $v =$ Constant $\\$ Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.$\\$ b. Take the relation:$\\$ $v=\sqrt{\dfrac{\gamma P}{\rho}} ...(i)$$\\$ For one mole of an ideal gas, the gas equation can be written as:$\\$ $PV = RT$$\\$ $P=\dfrac{RT}{V} ....(ii)$$\\$ Substituting equation (ii) in equation (i), we get:$\\$ $v=\sqrt{\dfrac{\gamma RT}{V \rho}}\\ =\sqrt{\dfrac{\gamma RT}{M}} ....(iv)$$\\$ Where, $\\$ Mass,$M=\rho V $ is a constant$\\$ $\gamma $ and $R$ are also constants$\\$ We conclude from equation (iv) that $v \propto T .$$\\$ Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.$\\$

5   You have learnt that a travelling wave in one dimension is represented by a functiony $= f ( x , t )$ where $x$ and $t$ must appear in the combination $x-vt$ or $ x+vt,$ i.e. $ y=f(x\pm vt).$ Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:$\\$ (a)$(x-vt)^2$$\\$ (b)$ \log [\dfrac{x+vt}{x_0}]$$\\$ (c)$\dfrac{1}{x+vt}$

Solution :

Explanation:$\\$ (a) No. For $x = 0$ and $t = 0,$ the function $(x - vt )^2$ becomes $0.$ Hence, for $x = 0$ and $t = 0$, the function represents a point and not a wave. (b) Yes. For $x = 0$ and $t = 0$, the function $\\$ $\log (\dfrac{x+vt}{x_0})=\log 0=\infty$$\\$ Since the function does not converge to a finite value for x = 0 and t = 0, it represents a travelling wave.$\\$ (c) No. For $x = 0$ and $t = 0$, the function$\\$ $\dfrac{1}{x+vt}=\dfrac{1}{0}=\infty$$\\$ Since the function does not converge to a finite value for $x = 0$ and $t = 0,$ it does not represent a travelling wave.$\\$ 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.