# Waves

## Class 11 NCERT Physics

### NCERT

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 $340 m s^{-1} ? ( g = 9. 8 m s^{-2} )$

$3$ 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.