Waves

Waves Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

standard equation of a progressive wave is given by

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

This is travelling along positive x- direction

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

Comparing with standard equation

(a) amplitude =5m

(b) k=2 $\frac{π}{γ}$ =0.4x

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

(c) w=10 $π$

w=2 $π v = 100 π$

frequency v= 100 $π / 2 π$ =50Hz

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

=100 $π / 0.4 π$ =1000/4

250m/s

(e) y= 5sin(100 $π t - 0.4 π x$ )

dy/dt = particle velocity

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

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

so particle velocity amplitude =(dy/dt)_max=5(100 $π$ ) $* 1$ =550 $π$ m/s

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4 months ago

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.

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Waves Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(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

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4 months ago

If c is r.m.s. 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.

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Waves Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

we know that rms speed of molecules of a gas

C= $\sqrt[]{\frac{3 p}{ρ}} = \sqrt[]{\frac{3 R T}{M}}$

Where M is the molar mass of the gas

Speed of sound wave in gas v= $\sqrt[]{\frac{γ p}{ρ}} = \sqrt[]{\frac{γ R T}{M}}$

On dividing above equation we get

c/v = $\sqrt[]{\frac{3 R T}{M} * \frac{M}{γ R T}}$

c/v = $\sqrt[]{\frac{3}{γ}}$

where $γ$ = adiabatic constant for diatomic gas

$γ = \frac{7}{5}$

c/v=constant

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4 months ago

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 km s^-1 in solid parts and of 5 km 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?

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Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Speed of wave in solid =8km/s

Speed of wave in liquid = 5km/s

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

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

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

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

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5 months ago

15.27 A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

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Contributor-Level 10

Ultrasonic beep frequency emitted by bat, $ν$ = 40 kHz

Velocity of the bat, $v_{b}$ = 0.03v, where v = velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

$ν^{'} = (\frac{v}{v - v_{b}}$ ) $ν$ = $(\frac{v}{v - 0.03 v}$ ) $* 40$ = $\frac{40}{0.97}$ kHz = 41.24 kHz

This frequency is reflected by the stationary wall ( $v_{s} = 0)$ towards the bat

The frequency ( $ν'') o f t h e r e c e i v e d s o u n d i s g i v e n b y t h e r e l a t i o n :$

$ν^{''} = (\frac{v + v_{b}}{v}$ ) $ν'$ = ( $\frac{v + 0.03 v}{v}$ ) $* 41.24 k H z$ = 1.03 $* 41.24$ = 42.47 kHz

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5 months ago

15.26 Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s^–1, and that of P wave is 8.0 km s^–1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur ?

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Contributor-Level 10

Let $v_{s}$ and $v_{p}$ be the velocities and $t_{s}$ and $t_{p}$ be the time taken to reach the seismograph from the epicentre of S and P waves respectively.

Let L be the distance between the epicentre and the seismograph.

We have:

L = $v_{s} t_{s}$ …. (i)

L = $v_{p} t_{p}$ …. (ii)

It is given, $v_{s}$ = 4 km/s and $v_{p}$ = 8 km/s

From equation (i) and (ii), we get

4 $t_{s}$ = 8 $t_{p}$ or $t_{s} =$ 2 $t_{p}$ …. (iii)

It is also given,

$t_{s} - t_{p} = 240 s$ so $t_{p} = 240 s a n d t_{s} = 480 s$

From equation (ii), we get, L = 8 $* 240$ = 1920 km

Hence, the earthquake occurred at a distance of 1920 km from the seismograph.

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5 months ago

15.25 A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^–1. What is the frequency of sound reflected by the submarine ? Take the speed of sound in water to be 1450 m s^–1

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Contributor-Level 10

Operating frequency of the SONAR system, $ν$ = 40 kHz

Speed of enemy submarine, $v_{e} =$ 360 km/h = 100 m/s

Speed of sound in water, v = 1450 m/s

The source is at rest and the observer (enemy submarine) is moving towards it. Hence, the apparent frequency (f') received and reflected by the submarine is given by the relation:

f = ( $\frac{v + v_{e}}{v}) ν =$ ( $\frac{1450 + 100}{1450}) * 40$ = 42.76 kHz

The frequency (f') received by the enemy submarine is given by the relation:

f' = ( $\frac{v}{v - v_{e}}$ )f = ( $\frac{1450}{1450 - 100}$ ) $* 42.76$ = 45.93 kHz

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5 months ago

15.24 One end of a long string of linear mass density 8.0 * 10^–3 kg m^–1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.

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Contributor-Level 10

The equation of a travelling wave propagating along the positive y- direction is given by the displacement equation:

y (x, t) = a sin ( $ω t - k x)$ ……… (i)

Linear mass density $μ =$ 8.0 $* 10^{- 3}$ kg/m and frequency of the tuning fork, $ν$ = 256 Hz

Amplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)

Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 $* 9.8$ = 882 N

The velocity of the transverse wave, v is given by the relation:

v = $\sqrt{\frac{T}{μ}}$ = $\sqrt{\frac{882}{8.0 * 10^{- 3}}}$ = 332 m/s

Angular frequency, $ω$ = 2 $π ν$ = 2 $π * 256$ = 1608.5 rad/s = 1.6 $* 10^{3}$ rad/s

Wavelength, $λ$ = $\frac{v}{ν}$ = $\frac{332}{256}$ &

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The equation of a travelling wave propagating along the positive y- direction is given by the displacement equation:

y (x, t) = a sin ( $ω t - k x)$ ……… (i)

Linear mass density $μ =$ 8.0 $* 10^{- 3}$ kg/m and frequency of the tuning fork, $ν$ = 256 Hz

Amplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)

Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 $* 9.8$ = 882 N

The velocity of the transverse wave, v is given by the relation:

v = $\sqrt{\frac{T}{μ}}$ = $\sqrt{\frac{882}{8.0 * 10^{- 3}}}$ = 332 m/s

Angular frequency, $ω$ = 2 $π ν$ = 2 $π * 256$ = 1608.5 rad/s = 1.6 $* 10^{3}$ rad/s

Wavelength, $λ$ = $\frac{v}{ν}$ = $\frac{332}{256}$ = 1.297 m

Propagation constant, k = $\frac{2 π}{λ}$ = $\frac{2 π}{1.297}$ = 4.844 /m

Substituting these values in the displacement equation, we get

y (x, t) = a sin ( $ω t - k x)$

y (x, t) = 0.05 sin (1.6 $* 10^{3}$ t – 4.844 x) m

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15.23 A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) Frequency (ii) Wavelength (iii) Speed of propagation? (b) If the pulse rate is 1 after every 20 s (that is the whistle is blown for a split of second after every 20 s) is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz ?

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The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed in that medium. The short pip produced after 20 s does not mean that the frequency of the whistle is $\frac{1}{20} o r$ 0.05 Hz. It means that 0.05 Hz is the frequency of repetition of the pip of the whistle. So the answers are

(a)

(i) No

(ii) No

(iii) Yes

(b) No

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5 months ago

15.22 A travelling harmonic wave on a string is described by

y(x, t) = 7.5 sin (0.0050x +12t + $π$ /4)

(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?

(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.

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