Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen: Overview, Questions, Preparation

Explanation- Speed of wave in solid =8km/s

Speed of wave in liquid = 5km/s

Required time = [ $\frac{1000-0}{8}+\frac{3500-1000}{5}+\frac{6400-3500}{8}$ ] $\times 2$

=[ $\frac{1000}{8}+\frac{2500}{5}+\frac{2900}{8}$ ] $\times 2$

=[125+500+362.5] $\times 2$ =1975       (diameter = radius $\times 2$ )

As we are considering at diametrically opposite point, hence there is a multiplication of 2

Explanation- We know that rms speed of molecules of a gas

C= $\sqrt[]{\frac{3p}{\rho }}=\sqrt[]{\frac{3RT}{M}}$

Where M is the molar mass of the gas

Speed of sound wave in gas v= $\sqrt[]{\frac{\gamma p}{\rho }}=\sqrt[]{\frac{\gamma RT}{M}}$

On dividing above equation we get

c/v = $\sqrt[]{\frac{3RT}{M}\times \frac{M}{\gamma RT}}$

c/v = $\sqrt[]{\frac{3}{\gamma }}$

where $\gamma$ = adiabatic constant for diatomic gas

$\gamma =\frac{7}{5}$

c/v=constant

15.3 Given below are some functions of x and t to represent the displacement of an elastic wave.

(a) y = 5 cos (4x ) sin (20t)

(b) y = 4 sin (5x – t/2) + 3 cos (5x – t/2)

(c) y = 10 cos [(252 – 250) πt ] cos [(252+250)πt ]

(d) y = 100 cos (100πt + 0.5x )

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.

Explanation- (a) the equation y = 100 cos (100πt + 0.5x ) is representing a travelling wave along x-direction

(b) the equation y = 5 cos (4x ) sin (20t) represents a stationary wave , because it contains sin cos terms ., combinations of two progressive waves.

(c) the equation y = 10 cos [(252 – 250) πt ] cos [(252+250)πt ] involving sum and difference of two near by frequencies 252 and 250 have this equation represents beats formation.

(d) as the equation y = y = 4 sin (5x – t/2) + 3 cos (5x – t/2) involves negative sign with x , have if represents a travelling wave along x – direction.

15.4 In the given progressive wave y = 5 sin (100πt – 0.4πx ) where y and x are in m, t is in s. What is the

(a) Amplitude

(b) Wave length

(c) Frequency

(d) Wave velocity

(e) Particle velocity amplitude.

Explanation- standard equation of a progressive wave is given by

Y=asin(wt-kx+ $\varnothing$ )

 This is travelling along positive x- direction

Given equation is y=5sin(100 $\pi t-0.4\pi x$ )

Comparing with standard equation

(a) amplitude =5m

(b) k=2 $\frac{\pi }{\gamma }$ =0.4x

wavelength =2 $\frac{\pi }{k}$ = $\frac{2\pi }{0.4\pi }=\frac{20}{4}=5m$

(c) w=10 $\pi$

w=2 $\pi v=100\pi$

frequency v= 100 $\pi /2\pi$ =50Hz

(d) wave velocity $v=\frac{w}{k}\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{k}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{w}\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{k}=2\mathrm{\pi }/\mathrm{\lambda }$

=100 $\pi /0.4\pi$ =1000/4

250m/s

(e) y= 5sin(100 $\pi t-0.4\pi x$ )

dy/dt = particle velocity

dy/dt = 5(100 $\pi )$ cos[100 $\pi t-0.4\pi x$ ]

for particle velocity amplitude (dy/dt)_max which will be for cos[100 $\pi t-0.4\pi x$ ]_max=1

so particle velocity amplitude =(dy/dt)_max =5(100 $\pi$ ) $\times 1$ =550 $\pi$  m/s