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Prof. Muhammad Amin 1

Chapter-8 waves

Type equation here.

Q. 8.1 (a) Describe Newton’s formula for the speed of sound and explain it was corrected

by Laplace?

(b) What is the effect of pressure, density and temperature on speed of sound in air

(gas)?

Ans. (a) Newton’s formula for the speed of sound in Air

The speed of sound waves depends on

The compressibility elasticity E

And inertia i.e. density of the medium through which they are travelling.

Newton proved the formula for the speed of sound in fluids (i.e. liquid, gas) which is

given by

EV ____________(1)

For fluids, the elasticity is measure by elastic modulus.

Elastic modulus stress

=strain

__________________(2)

Stress means a change of pressure. Strain is the ratio of change in volume to the

original volume. (i.e. ΔVV

)

Assumption: - Newton supposed that sound waves travel through air under isothermal

process, i.e. a process in which temperature remains constant. Therefore we can apply

Boyle’s law.

Let V is the volume of air at pressure P. If pressure is increased from P to (P P) at

constant temperature, then the volume decreases from V to ( V - V). Thus according

to Boyle’s law, we can write as

PV P ΔP V ΔV

PV PV PΔV+ΔPV ΔPΔV

PV PV = PΔV+ΔPV ΔPΔV

0 PΔV+ΔPV ΔPΔV__________(3)

If the increase in pressure P is very small then the decrease in volume V will also be

very small, therefore, PV is very small so it can be neglected

0 PΔV+ΔPV

or PΔV ΔPV

ΔPVor P

ΔV

Which may be written as

Waves Chapter

8

Important long questions


Chapter-8 waves

ΔP stressP = = = E

ΔV strain

V

Putting this vale of E in equation (1)

PV = _____________(4)

For air at normal pressure

P = 1013961.6 dyne/cm2

= 0.001293 gm/cm3

Putting in equation (4)

1013961.6V =

0.001293

V = 784193039.4

V = 2800344cms

or V = 280ms

1

1

But the experimental value of speed of sound in air is 33200 cms –1 (or 332 ms–1). The

error was nearly 16%.

This error was removed by Laplace.

Laplace’s Correction

Laplace said that during the propagation of sound in air compression and rarefaction

occurs so rapidly that there is no time to exchange heat ; the temperature does not

remain constant. Therefore, propagation of sound through air is not an isothermal

process.

Laplace, assumption

He assumed that propagation of sound is an adiabatic process. Thus the relation

between P and V, for adiabatic process is given by

PV = constant __________(5)

Where is a constant and its value depends upon the nature of the gas.

P

V

molar specific heat at constant pressure C= =

molar specific heat at constant volume C

Consider a gas having pressure P and volume V. If its pressure is increased by a small

amount P and volume is decreased by a small amount P.

Hence

PV P+ΔP V ΔV

V

PV P+ΔP V ΔV Multiplying and dividing by V.V

PV P+ΔP VΔV

1 __________( )V

6

Applying binomial theorem and neglecting the higher value ofΔV

V

, then


Chapter-8 waves

ΔV Δ

V

V

V

1 1 putting in equation (6)

ΔV

P = P + ΔPV

or P

1

= PΔV ΔV

P ΔP ΔPV V

ΔV ΔVP ΔP ΔP

V V

0

As ΔV

ΔPV

is small because V < < V, so it can be neglected.

ΔVP + ΔP

VΔV

or P = ΔPVΔP stress

or P= = = EΔV strain

V

0

Putting E = P in equation (1)

Laplace’s corrected formula for the speed of sound in air is

PV ____________( )7

For air 21.42 P= 1013961.6 dyne/cm = 0.001293 gm/cm3

1.42×1013961.6 1439825.47V = =

0.001293 0.001293

V = 1113554116

V = 33369.95cms

or V = 333ms

1

1

This value is very close to the experimental value.

(b) Effect of pressure, density and temperature

1. Effect of Pressure

As speed of sound through air is given by

PV

Since density is directly proportional to the pressure. With the increase or decrease of

pressure the density of air also increases or decreases.

Thus the ratio of

P remains constant. Therefore the speed of sound does not change

with the change of pressure.

2. Effect of Density

At the same temperature and pressure for the gases having the same value of , the

velocity is inversely proportional to the square root of their density.


Chapter-8 waves

P

V =

e.g. The speed of sound in hydrogen is four times its speed in oxygen as density of

oxygen is 16 times that of hydrogen.

3. Effect of Temperature

When a gas is heated at constant pressure, its volume is increased and its density is

decreased.

As

PV =

So, the speed is increased the rise in temperature.

Q.8.2 Show that one degree Celsius rise in temperature produces approximately 0.61

increase in speed of sound?

Ans. When a gas is heated at constant pressure, its volume is increased and its density is

decreased.

As

PV =

So, the speed is increased the rise in temperature.

LetV0 Speed of sound at 0oC

Vt Speed of sound at 0oC

o = Density of gas at 0oC

t = Density of gas at 0oC

Then

o

o

PV = _________( )1

And

t

t

PV = _________( )2

Dividing equation (2) by (1)

t t

o

t

P

V=

PV

t

o t

V= ________________( )

Vo 3

We have a relation for volume expansion of a gas.

___________( )t oV V t1 4

Where Vo = Initial volume of a gas at 0 oC

Vt = final volume of a gas at t oC

t = temperatute in oC

= coefficient of volume expansion of the gas.


Chapter-8 waves

For all gases, 1

273Putting is equation (4)

_____________( )t oV V t

11 5

273

MassDensity

Volume

,o t

o t

m m

V V

,o t

o t

m mor V V

Putting these values in equation (5)

m

t

o m

o

t1

273

t o

t1 11

273

o t

tor 1

273

Putting the value of o in equation (3)

tt

o t

t

V

V

1273

___________( )t

o

V t

V 1 6

273

t

o

V t

V

273

273

ot C T273

ooO C T273

________________( )t

o o

V T

V T 7

Relation between velocity and temperature.

Thus, the speed of sound is directly proportional to the square root of absolute temperature.

Expanding R.H.S of equation (6) by using Binomial theorem.


Chapter-8 waves

t

o

v t t

v

1

2

1 1273 273

......t

1

12 273

t o

tor v v 1

546

't o o

tV V V

546

t oV V t 332

546

. ________________( )t ov v t0 61 8

This shows that one degree Celsius (or centigrade) rise in temperature produces

approximately 0.61cms–1 increase in the speed of sound.

Q. 8.3 (a) Discuss different modes of vibration of transverse stationary waves in a

stretched string? OR

Describe the stationary waves produced in a stretched string. Prove that the

frequency of overtones in a stretched string are simple multiples of fundamental

frequency i.e. fn = n f1

(b) What is meant by fundamental frequency and overtone?

Ans. (a) Consider a string of length which is kept stretched by clamping its ends so that the

tension in the string is F.

If we pluck the string at its middle point, two transverse waves will start from that

point. They move in opposite direction and are reflected back from the clamped ends

superpose and form the stationary waves.

As the two ends of the string are clamped, no motion will take place there so nodes will

be formed at the two ends.

1. First Mode of Vibration

If we pluck the string at its middle point, the string vibrates in one loop as shown in

fig (a)

As the two ends of the string are at rest, they form two nodes and one antinode

between them.

Let frequency = f1

Wavelength = 1

Length of the string =

Speed of the wave = v


Chapter-8 waves

Distance between two nodes 1λ=2

1

1

λ=

2λ ___________(1)

2

The speed V of the waves in the string depends upon the tension in the string

depends F of the string and m, the mass per unit length of the string. The speed is

given by

FV = __________(2)

m

As 1 1

1

VV = f λ or =

λf1

Using equation no.(1)

V_______________(3)

2f 1

Putting the value of V from equation no.(2)

F

m2

1 F___________(4)

2 m

f

f

1

1

2. Second Mode of Vibration

Now let the same string is plucked from one quarter 1

th4

of its length and it

vibrates in two loops with another frequency f2 as shown in fig (b).

If 2 is the wavelength, then the distance between three nodes is

2 2

2

2

λ λ+ =

2 22λ

=2λ =

Velocity of the waves remain constant because it depends upon m and F

2 2V = λ f Putting the value of 2

2 2VV orf f

Multiplying and dividing by 2


Chapter-8 waves

V= 2

2

_____________( )

f

Vf

f f

2

1

2 1

22 5

3. Third Mode of Vibration

If the same string is plucked at one sixth 1

i.e. th6

of its length it vibrates in three

loops with another frequency f3 as shown in fig (c).

If 3 is the wavelength then the distance between four nodes is

3 3 3

3 3 3

3

λ λ λ

2 2 2λ λ λ

λ3

2

2

Or 3

2λ =

3

3 3V = λ f

Or 3

V

λf 3

Or

V

2f

33

Or V

3 ×2

f 3

V

=2

f

1

3f f3 1

This is called third harmonic

Similarly f4 = 4f1

f5 = 5f1

- - - - - -

- - - - - -

fn = nf1

Where n = number of loops and its value is n = 1, 2,3, ----------


Chapter-8 waves

(b) Fundamental Frequency

The lowest frequency f1 with which transverse wave is produced in a string is called

fundamental frequency.

Harmonic or Overtone

The frequencies other than the fundamental frequency, are called harmonics or

overtones.

Q. 8.4 Describe the stationary waves in air columns?

Ans. Stationary Waves in Air Columns

A wave produced by the simultaneous transmission of two similar. Wave motion in

opposite directions called stationary wave.

Stationary waves are caused interference between waves of the same frequency in such

a way that the combined intensity varies between maximum and minimum over the

region of interference.e.g. A vibrating air column in the organ pipe.

The relationship between the incident wave and the reflected wave depends on whether

the reflecting end of the pipe is open or closed. If the reflecting end is open the air

molecules have complete freedom of motion and this behaves as an anti -node. If the

end is closed, then it behaves as a node because the movement of the molecules is

restricted.

Open organ pipe

The mode of vibration of an air column in pipe open at both ends are shown in fig. The

longitudinal waves produce in the pipe have been represented by transverse curved lines

showing the varying amplitude of vibration of the air particles at points along the axis of the

pipe.


Chapter-8 waves

The frequency of any harmonic is given by

V

2nf n ___________(1)

Where n = 1, 2, 3---------

V is the velocity of sound in air and is the length of the pipe equation (1) can be

written as

fn = nf1_____________(2)

Closed organ pipe

If the pipe is closed at one end and open at the other the

closed end is a node.

The nodes of vibration in this case are shown in figure. In

case of fundamental note the distance between a node and

antinode is one fourth of the wavelength.

1λi.e. =4

Or 1λ = 4

1

1

VV = λ =

λf or f 1 1 Putting the value of 1

___________( )V

f 1 34

It can be proved that in a pipe closed at one end only odd

harmonics are generated which can be given by equation.

V= n ___________(4)

4nf

Where n = 1, 3, 5, --------

Q. 8.11 (a) Explain Doppler’s effect. Discuss its four cases?

(b) Describe the applications of Doppler’s effect?

Ans. (a) Doppler’s Effect

The apparent change in frequency (or pitch of sound) due to relative motion between

the source of sound waves and observer, is called Doppler Effect. This effect was

observed by Johan Doppler when he was observing the frequency of light emitted from

distant starts.

When an observer is standing on a railway platform the pitch of the whistle of an

approaching engine is heard to be higher. But when the engine moves away the pitch of

the whistle becomes lower.

There are four possible cases to discuss the Doppler effect.

Case (1)

When Observer Moves Towards the Stationary Source of Sound

The change in frequency due to Doppler effect can be calculated easily if the relative

motion between the source and the observer is along a straight line joining them.


Chapter-8 waves

Suppose V is the velocity of sound in the medium and the source emits a sound of

frequency f and wavelength.

If both the source and the observer are stationary then

__________(1)

__________(2)

V= _________(3)λ

V f

V

f

or f

If the observer A moves towards the source with a velocity uo as shown in fig.

The relative velocity of the waves and the observer is increased to (V = u o).

Then the number in one second or changed frequency fA is given by

_________(4)oA

V uf

Putting the value of from equation (2).

oA

V uf

V

f

Or V

Vo

A

uf f

As oV u

V

1

Af f

So the pitch of sound increases.

Case (2)

When Observer Moves away from the Stationary Source of Sound

Suppose the observer moves away from the stationary source with the same velocity u o,

and then he finds that the resultant velocity (i.e. relative velocity) of the waves and

observer is (V – uo). Then the modified frequency i.e. changed frequency which the

observer heard is fB

oB

V uf f

V

As oV u

V

1

Bf f

So the pitch of the sound decreases.


Chapter-8 waves

Case (3) When Source of Sound Moves towards the Stationary Observer

Suppose the source of sound moves towards the stationary observer with velocity u s as

shown in fig.

Then in one second, the waves are compressed by an amount known as Doppler shift

represented by .

su f

Or su

f ___________(1)

The compression of waves is due to the fact that same number of waves is contained in

a shorter space depending upon the velocity of the source. The wavelength for observer

C is then

C

sC

sC

uV

f f

V u

f

The modified frequency for observer C is

Cputting the value of λC

C

Vf

C

s

Vf

v u

f

C

s

Vf f

V u

, C

s

Vf f

V u

1

So the pitch of sound increases

Case (4) When Source of Sound Moves away from the Stationary Observer

Suppose the source f sound moves away from the stationary observer with velocity u s.

The modified frequency for observer D will be

D

s

s

D

Vf f

V u

V

V u

f f

1

So the pitch of sound decreases.


Chapter-8 waves

(b) Applications of Doppler’s Effect

1. Doppler effect is used in measuring the speed of automobile by traffic police. For

this a radar gun is fixed on police car. The frequency change is measured using

beats and hence the speed of the automobile is determined.

2. We can find the motion of an object under water. When sound waves (called

sonar) under water are reflected from moving submarine the change in frequency

of the reflected waves help to find the speed and position of the submarine.

3. The waves sent out from radar are reflected from the aeroplanes. The frequency of

the reflected waves is decreased if the plane is moving away from the radar.

The frequency increases if the plane is moving towards the radar.

4. A bat finds the position and nature of objects by sending ultrasonic waves. The

working of a bat is based upon Doppler’s effect.

5. The velocities of earth satellites are found out by the change in frequency of radio -

waves emitted by the satellites.


Chapter-8 waves

8.1 What features do longitudinal waves have in common with transverse waves?

Ans. 1. In both types of waves the particles vibrate about their mean position.

2. Both types of waves transfer energy from one place to another.

3. For both types of waves the product of frequency and wave length is equal to the

velocity of the wave.i.e. v = f

4. Both types of waves can be setup in solids.

5. Both types of waves produce disturbance in the medium though which they travel.

8.2 Is it possible for two identical waves travelling in the same direction along a string

to give rise to a stationary wave?

Ans. It is not possible.

Reason:-

Stationary waves are only produced when two identical waves travelling in the same

line but in opposite direction.

8.3 A wave is produced along a stretched string but some of its particles permanently

show zero displacement. What type of wave is it?

Ans. This is stationary wave.

Reason:-

In stationary waves the particles at nodes are permanently at rest so show zero

displacement and some particles show maximum displacement called anti-nodes.

8.4 Explain the terms crest through node and antinode?

Ans.

Trough:-The lowest position below the mean line of a transverse wave is called trough.

Crest:-The highest position above the mean line of a transverse wave is called crest.

Node:-The point where the particle of medium vibrates with minimum or zero

amplitude is called a node.

Anti-node:-The point where the particle of medium vibrates with maximum amplitude

is called antinodes.

\

Important short questions


Chapter-8 waves

8.5 Why does sound travel faster in solids than in gases?

Ans. The speed of sound in a medium of elasticity E and density is given by

EV

𝜌solid > 𝜌gases

But E solid >> E gases

As the density of solid is larger than gases but modulus of elasticity of solid is much

larger than gases. Therefore sound travels faster in solid than gases.

8.6 How are beats useful in tuning musical instruments?

Ans. The number of beats produced per second is equal to the difference between the

frequencies of two tuning forks. If we know the frequency of the other can be

calculated by counting beats. Therefore beats are useful for tuning a musical

instrument.

One can tune a string instrument by tightening or loosening the string. When no beat

are heard the instrument is said to be tuned.

8.7 As a result of a distant explosion, an observer senses a ground tremor and then

hears the explosion. Explain the time difference?

Ans. The speed of sound in a medium of elasticity E and density is given by

EV

Due to explosion disturbance are produced. These follow two different paths, one

through the earth and the other through atmosphere. Because sound travels faster in

solids than in gases. So sound travelling through air will be heard later.

8.8 Explain why sound travels faster in warm air than in cold air?

Ans. The speed of sound in gas is given by

PV

In warm air the temperature of air is increased the pressure also increases and hence

density decreases. So the speed of sound increases. On the other hand for cold air the

temperature is small.

With the decreases of temperature, the pressure decreases and so density increases.

Thus the speed of sound decreases. That is why the sound travels faster in warm air

than in cold air.

8.9 How should a sound source move with respect to an observer so that the frequency

of its sound does not change?

Ans. When sound source moves perpendicular to the line joining the source and observer

then there will be no change in its frequency of sound.


Chapter-8 waves

OR

If relative velocity of sound source and observer is zero. Hence the apparent frequency

of sound will not be changed.

Example

If observer is moving on a circular path such that source of

sound is stationary at the center of the circle then there will be

no apparent change of frequency.

8.10 Prove that speed of sound in hydrogen is 4-times as that of speed of sound in

oxygen?

Ans p

V

1

V

As 𝜌𝑜𝑥𝑦𝑔𝑒𝑛 = 16𝜌ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛

h

O

V 116 4

1V

16

Vh = 4Vo

Hence speed of sound in hydrogen is 4 times its speed in oxygen.

8.11 Why Radar cannot detect under water objects?

Ans Radar cannot detect under water objects

Reason:-

Water is a denser medium having high refractive index compared to air. Due to this

fact when radar waves are focused into water then they can only penetrate few cm into

water due to their low wavelength.

8.12 A wave has speed 400 m / sec. Find wavelength of a wave if frequency is 2 kHz.

Ans v = 400 m / s

λ = ?

32 2 10f kHz Hz

V = f λ

V

f

3

400

2 10

= 0. 2 m

8.13 The speed of sound in air at 0oC is 332 ms-1. Find its speed at 20oC.

Ans The speed of sound varies directly proportional to temperature and can be measured by

using the following equation.

T oV V 0.61T

VT = 332 + 0.61 (20) 1

oV 332ms

VT = 344.2ms-1


Chapter-8 waves

8.14 What are the classification of waves?

Ans Waves can be classified into three groups.

(i) Mechanical Waves

Such waves which required certain medium for their propagation are called mechanical

waves. E.g water waves, sound waves

(ii) Electromagnetic Waves

Such waves which require no medium for their propagation. Their range is from low

frequency radio waves to high frequency gamma and cosmic rays.

e.g Light waves, radio waves

(iii) Matter Waves

Wave associated with particles are called matter waves.

E.g Waves associated with fast moving electrons, protons

8.15 Write characteristics of stationary waves?

Ans 1.Distance between two consecutive nodes or anti nodes is equal to2

2.Distance between node and next anti node is 4

3.In stationary waves, Nodes always remain at rest, so energy cannot flow past

through nodes

4.Amplitude is maximum at antinodes and minimum at nodes.

5.Points of constructive interference are called antinodes while points of de structive

interference are called nodes.

8.16 State and explain the principle of super position?

Ans Principle of Superposition

If a particle of the medium is simultaneously acted upon by two waves then the

resultant displacement of the particle is the algebraic sum of their individual

displacements. This is called principle of superposition. \

Explanation

If y1, y2, y3, ---------- yn are the displacements caused by the individual waves and y is

the resultant displacement then

y = y1 + y2 + ---------+yn

8.17 What are condition of constructive interference and destructive interference?

Ans Constructive Interference

The two waves interfere constructively, if the crest of one wave fall on the crest of the

other wave and trough of one wave fall on the tough of other wave respectively.

This condition is satisfied only if the path difference between than is integral multiple

of wave length. That is path difference n λ (where n = 0,1,2,3,……)

Destructive Interference

The two waves are paid to interfere destructively if crest of one wave fall on the trough

of the other wave and trough of one wave fall on the crest of the other wave

respectively.


Chapter-8 waves

This condition is satisfied only if the path difference between them is odd integral

multiple of 2

path difference (2n 1) 2

where n = 0,1,2,3,……

8.18 Differentiate b/w transverse and longitudinal waves?

Ans There are two kinds of progressive waves.

(i) Transverse Waves (ii) Longitudinal Waves

Transverse Waves

The waves in which the particles of the medium vibrate along a line perpendicular to the

direction of propagation of waves are known as transverse waves .e.g. Water waves, waves

along a string.

Longitudinal Waves

Waves in which particles of the medium vibrate along the line of propagation of wave are

called longitudinal or compressional waves. e.g. sound waves and waves along a compressed

spring.


Chapter-8 waves

8.1 The wavelength of the signals from a radio transmitter is 1500m and the frequency 200 kHz. What is the wavelength for a transmitter operating at 1000 kHz and with what speed the radio waves travel?

Solution:

Data: 1Wavelength of signals =λ =1500m

1Frequency of signals=f =200kHz

3

5

1

=200×10 Hz

f =2×10 Hz

2

6

2

Frequency for transmitter=f =1000kHz

f =10 Hz

To Find: Speed of RadioWaves = v=?

2Wavelength for transmitter=λ =?

Calculation: Using

1 1v=f λ

5

7

2 10 1500

30 10

8v=3×10

8v = 3×10 m / s Ans

Now 2 2v=f λ

2

2

vλ =

f

8

2

6

3 103 10

10

2λ = 300m Ans

8.2 A stationary wave is established in a string which is 120 cm long and fixed at both ends. The

string vibrates in four segments, at a frequency of 120Hz. Determine its wavelength and the

fundamental frequency?

Solution:

Data: Length of the string = = 120cm = 1.2 m

Number of loops = n = 4

Frequency of vibration = f4 = 120 Hz

To Find: Fundamental frequency = f1 = ?

Wavelength = 4 ?

Calculation:

For four loops

Numerical Problems


Chapter-8 waves

As string vibrates in four segments and distance between two consecutive nodes is 2

.

Therefore,

4 4 4 4

4

λ λ λ λ+ + +

2 2 2 2

λ4

2

4

4

2λ

4 2

1.2λ 0.6

2m

As, fn = nf1

f4 = 4f1

41

ff =

4

1

120f =

4

f1 = 30 Hz Ans

8.3 The frequency of the note emitted by a stretched string is 300 Hz. What will be the

frequency of this note when; (a) the length of the wave is reduced by one-third without

changing the tension. (b) the tension is increased by one-third without changing the length of

the wire.

Solution:

Let 1 be the wavelength of wave of frequency f1 = 300Hz

T = constant

Then v = constant

To find: (a) (When only wavelength of waves is reduced by one third) = f2 = ?

(b) (When only tension is increased by one third ) = 2f = ?

Calculation:

New wavelength = 12 1

λλ =λ -

3

1 13λ -λ=

3

12

2λλ =

3

Since v remains same if T is constant.

1 1f λ =v

And 2 2v f λ


Chapter-8 waves

1 1 2 2

11 1 2

21

f λ =f λ

2λf λ =f

3

2ff =

3

2 1

2

3f = f

2

3f = ×300

2

= 3×150

2f = 450Hz Ans

(b) Now 2f = ?

If tension is increased by 1

3

Now tension is T

= T = T +3

4T

T =3

In this problem

2

1 Tf =

2l m

2

1 4T/3f =

2l m

2

4 1 Tf =

3 2l m

2f = 1

2×f

3

2

2f = ×300

3

2

f = 346Hz Ans

8.4 An organ pipe has a length of 50cm. Find the frequency of its fundamental note and the next

harmonic when it is (a) open at both ends (b) closed at one end. (Speed of sound = 350 m/s)

Solution:

Data:

(a) Length of organ pipe=l=50cm=0.5m

Speed of sound = v=350m/s


Chapter-8 waves

To find: 1

2

Fundamental frequency=f =?When pipe is open at both ends.

Next harmonic frequency = f =?

(b) 1

3

Fundamental frequency ?

Next harmonic frequency ?

f

f

When pipe is closed at one end.

Calculation:

(a) When pipe is open at both ends.

1

1

v 350f

2 2 0.5

350350

1.0

f 350Hz

and 2 1f =2f =2×350

Ans2f = 700Hz

(b) When pipe is closed at one end.

1

1

v 350f = =

4l 4×0.5

350f = =175Hz

2

And 3 1f =3f

=3×175

3f = 525Hz Ans

8.5 A church organ consists of pipes, each open at one end, of different lengths. The minimum

length is 30mm and the longest is 4m. Calculate the frequency range of the fundamental

notes.

Solution:

Data: For minimum length

= 30mm

= 0.03 m

V = 340 m / s

For one end open (closed organ pipe fundamental frequency)

To find: f1 = ?

Calculation:

1

vf =

4l

340 340,00= =

4×0.03 4×3


Chapter-8 waves

1

34000=

12

f =2833Hz

For longest pipe

fi = ?

1 = 4m

1

1

1

vf =

4l

340 340= =

4×4 16

f =21Hz

So, the frequency range is 21 Hz to 2833Hz.

8.6 The absorption spectrum of faint galaxy is measured and wavelength of one of the lines identified as the Calcium α line is found to be 478 nm. The same line has a wavelength of 397 nm when measured in a laboratory. (a) Is the galaxy moving towards or away from the Earth? (b) Calculate the speed of the galaxy relative to Earth? (speed of light = 3.0 108ms-1

Solution:

Data: Apparent wavelength of calcium line = -9λ =478nm=478×10 m

Actual wavelength = -9λ=397nm=397×10 m To Find: C = v = 3×108 m/s Apparent frequency of light = f` = ? Calculation:

(a) Since c = f λ

8

-9

14

8

-9

c 3×10f = =

λ 397×10

f = 7.56×10 Hz

c 3×10and f = =

λ 478×10

=

14f 6.28×10 Hz Ans

Since apparent frequency f is less than actual frequency f

Galaxy is moving away from earth.

(b) Speed of galaxy relative to earth = vs = ?

Since galaxy moving away from observer

s

s

vf = f

v+v

v fv+v =

f

s

s

148

14

vfv = -v

f

fv -1 v

f

f= -1 c

f

7.56×10= -1 3×10

6.28×10

8

8

8

7

= 1.203-1 3×10

= 0.203 3×10

=0.609×10

= 6.09×10 m/s

7

sv = 6.1×10 m / s Ans

waves type equation here. 8 - profaminzcom.files.wordpress.com · cms ms 1 1 this value is very...

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