HSC PHQ 06 Superposition of Waves – Maharashtra Board Solutions

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06 Superposition of Waves

Multiple Choice Questions

1. In reflection of sound waves from rarer medium, there is phase change of…………………

(A)

(B)

(C)

(D)

Ans. (A)

2. In open organ pipe, first overtone produced is of such frequency that length of the pipe is equal to………………………..

(B)
(C)
(D)

Ans. (D)

3. In the equation of a simple harmonic progressive wave of wavelength ‘ ‘, the propagation constant is given by…………………

(A)
(B)
(C)
(D)

Ans. (A)

4. A stretched string of length vibrates in third overtone, the wavelength of stationary wave formed is………………..

(A)
(B)
(C)
(D)

Ans. (A)

5. In a simple harmonic progressive wave of amplitude , the maximum particle velocity is two times its wave velocity, then the wavelength of the wave is……………………
(A)
(B)
(C)
(D)

Ans. (C)

6. For two vibrating bodies to be in resonance, which of the following quantity should be equal?

(A) Wavelength
(B) Frequency

Ans. (B) Frequency

7. One beat means that the intensity of sound should be

(A) once maximum

(B) once minimum

(D) twice maximum and twice minimum

Ans. (C) once maximum and once minimum

8. When a transverse wave on a string is reflected from the free end, the phase change produced is………………….

(A) zero rad
(B)
(C)
(D)

Ans. (A) zero rad

9. Let and , be the two slightly different frequencies of two sound waves. The time interval between waxing and immediate next waning is………………….

[Mar 14]
(A)
(B)
(C)
(D)

Ans. (D)

10. A pipe open at both ends resonates to a frequency ‘ ‘ and a pipe closed at one end resonates to a frequency ‘ ‘. If they are joined to form a pipe closed at one end, then the fundamental frequency will be………………….

(A)
(B)
(C)
(D)

Ans. (A)

11. Let velocity of a sound wave be ‘ ‘ and ‘ ‘ be angular velocity. The propagation constant of the wave is…………………

[Oct 14]
(A)
(B)
(C)
(D)

Ans. (C)

12. The value of end correction for an open organ pipe of radius ‘ ‘ is……………..

(A)
(B)
(C)
(D)

Ans. (D)

13. When longitudinal wave is incident at the boundary of denser medium, then

(A) compression reflects as a compression.

(B) compression reflects as a rarefaction.

(D) longitudinal wave reflects as transverse wave.

Ans. (A) compression reflects as a compression.

14. The equation of a progressive wave is , where and are in cms and time in seconds. The maximum velocity of a particle is………………..

(A)
(B)
(C)
(D)

Ans. (A)

15. The fundamental frequency of transverse vibration of a stretched string of radius is proportional to………………….
(A)
(B)
(C)
(D)

Ans. (B)

16. A sonometer wire vibrates with frequency in air under suitable ‘load of specific gravity ‘ ‘. When the load is immersed in water, the frequency of vibration of wire will be………………..
(A)
(B)
(C)
(D)

Ans. (B)

17. If sound waves are reflected from surface of denser medium, there is phase change of……………………..

(A)
(B)
(C)
(D)

Ans. (D)

18. The working of RADAR is based on

(A) resonance

(B) speed of a star

(D) speed of rotation of sun

Ans. (C) Doppler effect

19. A sonometer wire vibrates with three nodes and two antinodes, the corresponding mode of vibration is……………….

(A) First overtone
(B) Second overtone
(C) Third overtone
(D) Fourth overtone

Ans. (A) First overtone

20. A sine wave of wavelength ‘ ‘ is travelling in a medium. What is the minimum distance between two particles of the medium which always have the same speed?

(A)
(B)
(C)
(D)

Ans. (B)

21. Velocity of transverse wave along a stretched string is proportional to ………………… tension in the string)

(A)
(B)
(C)
(D)

Ans. (A)

22. Doppler effect is not applicable when

(A) source and observer are at rest.

(B) there is a relative motion between source and observer.

(D) both are moving in same direction with different velocities.

Ans. (A) source and observer are at rest.

23. In stationary wave, the distance between a node and its adjacent antinode is…………………….

(A)
(B)
(C)
(D)

Ans. (B)

24. A wave is travelling in the negative direction of -axis, then its equation is …………………….
(A)

(B)

(C)

(D)

Ans. (B)

25. If the longitudinal wave travelling in rarer medium is incident on the boundary of denser medium, then the phase of wave changes by

(A)
(B)
(C)
(D)

Ans. (B)

26. A standing wave is produced on a string fixed at one end and free at other. The length of string must be an………………………

(A) odd integral multiple of

(B) integral multiple of

(D) integral multiple of

Ans. (A) odd integral multiple of

27. Two tuning forks have frequencies and respectively. On sounding these forks together, the time interval between two successive maximum intensities will be ………………….
(A)
(B)
(C)
(D)

Ans. (A)

28. Standing waves are produced on a string fixed at both ends. In this case

(A) all particles vibrate in phase

(B) all antinodes vibrate in phase

(D) all particles between two consecutive antinodes vibrate in phase

Ans. (C) all alternate antinodes vibrate in phase

Theory Questions

6.2 Progressive Wave

1. Express equation of one dimensional simple harmonic progressive wave travelling in the direction of positive -axis in ‘two’ different forms.

Ans:

i. Equation of simple harmonic progressive wave travelling along the positive -direction is given by,

where, amplitude of the wave,

ii. Different forms of equation of simple harmonic progressive wave:

a. Wave number,

b. Angular velocity,

c. Frequency,

d. Wave frequency,

[Any two formis]

2. State any two characteristics of progressive waves.

Ans: Characteristics of progressive waves:

i. Each particle in a medium executes the same type of vibration. Particles vibrate about their mean positions performing simple harmonic motion.

ii. All vibrating particles of the medium have the same amplitude, period and frequency:

iii. The phase (state of vibration of a particle), changes from one particle to another.

iv. No particle remains permanently at rest. Each particle comes to rest momentarily while at the extreme positions of vibration.

v. The particles attain maximum velocity when they pass through their mean positions.

vi. During the propagation of wave, energy is transferred along the wave. There is no transfer of matter. viii. Progressive waves are of two types -transverse waves and longitudinal waves.

ix. In a transverse wave, vibrations of particles are perpendicular to the direction of propagation of wave and produce crests and troughs in their medium of travel. In longitudinal wave, vibrations of particles produce compressions and rarefactions along the direction of propagation of the wave.

x. Both, the transverse as well as the longitudinal, mechanical waves can propagate through solids but only longitudinal waves can propagate through fluids.

[Any two characteristics]

6.3 Reflection of Waves

1. Explain the reflection of transverse and longitudinal waves from a denser medium and rarer medium.

Ans:

i. Reflection from a denser medium:

a. In case of a longitudinal wave, a compression is reflected back as a compression and a rarefaction is reflected back as a rarefaction.

b. In case of a transverse wave, a crest is reflected back as a trough and a trough is reflected as a crest.

ii. Reflection from a rarer medium:

a. In case of a longitudinal wave, a compression is reflected as a rarefaction and a rarefaction is reflected as a compression.

b. In case of transverse wave, a crest is reflected as a crest and a trough is reflected as a trough.

Superposition of Waves

1. State the principle of superposition of waves.

Ans: When two or more waves, travelling through a medium, pass through a common point, each wave produces its own displacement at that point, independent of the presence of the other wave. The resultant displacement at that point is equal to the vector sum of the displacements due to the individual wave at that point.

6.5 Stationary Waves

1. State the points of comparison between progressive waves and stationary waves.

Ans.

Sr.No.	Progressive waves	Stationary waves
i.	The disturbancetravels from oneregion to the otherwith definite velocity.	Disturbance remainsin the region where itis produced, velocityof the wave is zero.

ii.	Amplitudes of allparticles are same.	Amplitudesparticles are different.
iii.	Particles do not crosseach other.	All the particles crosstheir mean positionssimultaneously.
iv.	All the particles aremoving.	Particles at theposition of nodes arealways at rest.
vi.	There is no transferof energy.	Energy is transmittedfrom one region toanother.
particles are different.	All particles betweentwo consecutive nodesare moving in thesame direction and arein phase while those inadjacent loops aremoving in oppositedirections and differ inphase by 180.

2. Explain the formation of stationary waves by analytical method. Show that nodes and antinodes are equally spaced in a stationary wave.

Explain analytically how the stationary waves are formed. Hence show that the distance between node and adjacent antinode is .

Ans:

i. Consider two simple harmonic progressive waves of equal amplitudes (a) and wavelength propagating on a long uniform string in opposite directions.

ii. The equation of wave travelling along the -axis in the positive direction is given by,

The equation of wave travelling. along the -axis in the negative direction is given by,

iii. When these waves interfere, the resultant displacement of particles of string is given by the principle of superposition of waves as

iv. By using trigonometry formula,

v. Substituting in equation (1),

This is the equation of a stationary wave which gives resultant displacement due to two simple harmonic progressive waves.

vi. Condition for node:

Nodes are the points of minimum displacement.

This is possible if the amplitude is minimum, i.e., .

i.e., where

Distance between two successive nodes:

Distance between two successive nodes is .

vii. Condition for antinode:

Antinodes are the points of maximum displacement, i.e.,

i.e., where

Distance between two successive antinodes:

;

Distance between two successive antinodes is

viii. Nodes and antinodes are formed alternately and are equally spaced. Therefore, the distance between a node and an adjacent antinode is .

3. Explain the formation of stationary waves by analytical method. Show the formation of stationary wave diagrammatically. [Mar 17]

Ans:

i. Refer Subtopic 6.5: Q. No. 2.(i,ii,iii,iv,v).

ii.

Position and displacement of nodes and antinodes in a stationary wave

6.6 Free and Forced Vibrations

1. Distinguish between free vibrations and forced vibrations.

Ans:

Sr.No.	Free vibrations	Forced vibrations
i.	Free vibrations areproduced when abody is disturbedfrom its equilibriumposition and released.	Forced vibrations areproduced by anexternal periodic forceof any frequency.
ii.	To start freevibrations, the forceis required initiallyonly.	Continuous externalperiodic force isrequired. If externalperiodic force isstopped, then forcedvibrations also stop.
iii.	freeIs onncy.	The frvibratithe frextern
iv.	Energy of the bodyremains constant inthe absence offriction, air resistance,etc. Due to dampingforces, total energydecreases.	Energy of the body ismaintained constantby the externalperiodic force.
.	Amplitude ofvibrationswith time.	Amplitude is small butremains constant as longas external periodicforce acts on it.
vi.	Vibrationssooner or lopdepending on thedamping force.	Vibrations stop assoon as externalperiodic force isstopped.

[Any four differences]

2. Distinguish between free vibrations and forced vibrations (Two points).

Ans: Refer Subtopic 6.6: Q. No. 1 (Any Two Points)

6.7 Harmonics and Overtones

Draw a neat labelled diagram of the first two modes of vibrations of string stretched between two rigid supports.

Ans: First two modes:

(a) First mode

(b) Second mode

2. Draw neat labelled diagrams of the first two modes of vibrations of an air column open at both ends.

Ans:

i. For fundamental mode:

ii. For second mode

First overtone for vibrations of air column in a pipe open at both ends

With a neat labelled diagram, show that all harmonics are present in an air column contained in a pipe open at both the ends. Define end correction.

Ans:

i. For fundamental mode:

a. The fundamental tone or mode of vibrations of air column open at both ends is as shown in figure.

Refer Subtopic 6.7: Q. No. 2. (i) for figure

b. There are two antinodes at two open ends and one node between them.
c. Length of air column,

….[From equation (1)]

This is the fundamental frequency or the first harmonic and the lowest frequency of vibration.

ii. For second mode or first overtone:

a. The next possible mode of vibrations of air column open at both ends is as shown in figure. Refer Subtopic 6.7: Q. No. 2. (ii) for figure

b. Three antinodes and two nodes are formed.

c. Length of air column,

i.e.,

d. If and are frequency and wavelength of this mode of vibration of air column respectively, then

….[From equation (2)]

This is the frequency of second harmonic or first overtone.

iii. For third mode or second overtone:

a. In the next of vibrations of air column open at both ends, four antinodes and three nodes are formed.

Second overtone for vibrations of air column in a pipe open at both ends

b. Length of air column,

c. If and are the frequency and wavelength of this mode of vibration of air column respectively, then

….[From equation (3)]

This is the frequency of third harmonic or second overtone.

iv. Continuing in this manner, the frequency for overtone is,

where is the fundamental frequency and

v. Thus all harmonics are present as overtones in the modes of vibration of air column open at both ends.

vi. End correction: The distance between the open end of the pipe and the position of antinode is called the end correction.

With neat diagram explain fundamental mode of vibration of an air column in a pipe when,

i. Pipe is open at both ends.

ii. Pipe is closed at one end.

Hence, derive the expression for fundamental frequency in each case.

Ans:

i. For pipe open at both ends.

Refer Subtopic 6.7: Q. No. 3.(i).

ii. For pipe closed at one end :

a. The first mode of vibration of air column closed at one end is known as the fundamental mode.

Fundamental mode of vibrations of air column in a pipe closed at one end

b. Length of air column

where is the wavelength of fundamental mode of vibrations in air column.

c. If is the fundamental frequency, we have

….[From equation (1)]

..Considering end correction]

d. The fundamental frequency is also known as the first harmonic. It is the lowest frequency of vibration in air column in a pipe closed at one end.

Draw neat, labelled diagrams for the modes of vibration of a stretched string in second harmonic and third harmonic.

Ans:

i. For second harmonic:

Refer Subtopic 6.7: Q. No. 1. (b) for figure

ii. For third harmonic:

Show that only odd harmonics are present in an air column vibrating in a pipe closed at one end.

Ans:

i. For fundamental mode:

a. The first mode of vibration of air column closed at one end is known as the fundamental mode.

b. Length of air column

where is the wavelength of fundamental mode of vibrations in air column.

c. If is the fundamental frequency, we have

….[From equation (1)]

..Considering end correction] d. The fundamental frequency is also known as the first harmonic. It is the lowest frequency of vibration in air column in a pipe closed at one end.

ii. For second mode or first overtone:

a. The next mode of vibrations of air column closed at one end is as shown in figure.

b. Here the air column is made to vibrate in such a way that it contains a node at the closed end, an antinode at the open end with one more node and antinode in between.

c. If is the frequency and is the wavelength of wave in this mode of vibrations in air column, the length of the air column

d. The velocity in the second mode is given as

[From equation (2)]

e. This frequency is the third harmonic and the first overtone.

iii. For third mode or second overtone:

a. The next higher mode of vibrations of air column closed at one end is shown in the figure.

b. Here the same air column is made to vibrate in such a way that it contains a node at the closed end, an antinode at the open end with two more nodes and antinodes in between.
c. If is the frequency and is the wavelength of the wave in this mode of vibrations in air column, then length of air column,

d. The velocity this mode is given as

….[From equation (3)]

e. This frequency is the fifth harmonic and the second overtone.

iv. Continuing in a similar way, for the overtone the frequency is given as

Thus for a pipe closed at one end only odd harmonics are present and even harmonics are absent.

Show that all harmonics are present on a stretched string between two rigid supports.

Ans:

i. Fundamental mode:

a. If a string is stretched between two rigid supports and is plucked at its centre, the string vibrates as shown in figure.

b. It consists of an antinode formed at the centre and nodes at the two ends with one loop formed along its length.

c. If is the wavelength and is the length of the string, then

Length of loop

d. The frequency of vibrations of the string,

This is the lowest frequency with which the string can vibrate. It is the fundamental frequency of vibrations or the first harmonic.

ii. For second mode or first overtone:

a. For first overtone or second harmonic, two loops are formed in this mode of vibrations.

b. There is a node at the centre of the string and at its both ends.

c. If is wavelength of vibrations, the length of one loop

d. Thus, the frequency of vibrations is given as

Comparing with fundamental frequency,

Thus the frequency of the first overtone or second harmonic is equal to twice the fundamental frequency.

iii. For third mode or second overtone:

a. The string is made to vibrate in such a way that three loops are formed along the string as shown in figure.

b. If is the wavelength, the length of one loop is

c. Therefore the frequency of vibrations is

Comparing with fundamental frequency,

Thus frequency of second overtone or third harmonic is equal to thrice the fundamental frequency.

iv. Similarly for higher modes of vibrations of the string, the frequencies of vibrations are as , ..etc.

Thus all harmonics are present in case of a stretched string and the frequencies are given by, .

Discuss different modes of vibrations in an air column of a pipe open at both the ends. State the cause of end correction. Find the end correction for the pipe open at both the ends in fundamental mode. [July 17]

Ans:

i. Refer Subtopic 6.7: Q. No. 3.(i,ii,iii,iv,v).
ii. Cause of end correction:

a. When an air column vibrates either in a pipe closed at one end or open at both ends, boundary conditions demand that there is always an antinode at the open end(s) (since the particles of the medium are comparatively free) and a node at the closed end (since there is hardly any freedom for the particles to move).

b. The antinode is not formed exactly at the open end but it is slightly beyond the open end as air is more free to vibrate there in comparison to the air inside the pipe.

c. Also as air particles in the plane of open end of the pipe are not free to move in all directions, reflection takes place at the plane at small distance outside the pipe.

d. The distance between the open end of the pipe and the position of antinode is called the end correction.

iii. End correction for a pipe open at both ends:

, OR

What is meant by harmonics? Show that only odd harmonics are present as overtones in the case of an air column vibrating in a pipe closed at one end.

Ans:

i. Harmonics:

a. The term ‘harmonic’ is used when the frequency of a particular overtone is an integral multiple of the fundamental frequency.

b. In strings and air columns, the frequencies of overtones are integral multiples of the fundamental frequencies. Hence, they are termed as harmonics.

c. All harmonics may not be present in a given sound.

ii. Refer Subtopic 6.7: Q. No. 6.

Draw three neat diagram for modes of vibration of an air column in a pipe, when

i. the pipe is open at both ends, and

ii. the pipe is closed at one end.

Hence derive an expression for fundamental frequency in each case.

Ans:

i. Refer Subtopic 6.7: Q. No. 3.(i,ii,iii) diagrams only.

ii. Refer Subtopic 6.7: Q. No. 6. (i,ii,iii) diagrams only.

iii. Expression for fundamental frequency in a pipe open at both ends.

Refer Subtopic 6.7: Q. No. 3.(i).

iv. Expression for fundamental frequency in a pipe closed at one end.

Refer Subtopic 6.7: Q. No. 6.(i).

11. Show that all harmonics are present in case of an air column vibrating in a pipe open at both ends.

Show that even as well as odd harmonics are present as overtones in the case of an air column vibrating in a pipe open at both the ends.

Ans: Refer Subtopic 6.7: Q. No. 3.(i,ii,iii,iv,v)

What are harmonics and overtones (Two points)?

Ans:

i. Harmonics:

a. The term ‘harmonic’ is used when the frequency of a particular overtone is an integral multiple of the fundamental frequency.

b. The fundamental frequency (n) is considered as first harmonic and the frequency (2n) is the second harmonic which may or may not be present in a given sound. Hence, all harmonics may not be present in a given sound.

ii. Overtones:

a. The tones whose frequencies are greater than the fundamental frequency are called overtones.

b. The first frequency higher than the fundamental frequency is called the first overtone, the next frequency higher is the second overtone and so on. Overtones are only those multiples of fundamental frequency which are actually present in a given sound. Hence, all overtones are always present in a given sound.

6.8 Sonometer

Explain how law of length can be verified by using a sonometer.

Ans: Verification of first law:

i. By measuring length of wire and its mass, the mass per unit length (m) of wire is determined. Then the wire is stretched on the sonometer and the hanger is suspended from its free end.

ii. A suitable tension ( ) is applied to the wire by placing slotted weights on the hanger.

iii. The length of wire vibrating with the same frequency as that of the tuning fork is determined as follows.

iv. A light paper rider is placed on the wire midway between the bridges. The tuning fork is set into vibrations by striking on a rubber pad.

v. The stem of tuning fork is held in contact with the sonometer box. By changing distance between the bridges without disturbing paper rider, frequency of vibrations of wire is changed.

vi. When the frequency of vibrations of wire becomes exactly equal to the frequency of tuning fork, the wire vibrates with maximum amplitude and the paper rider is thrown off.
vii. In this way a set of tuning forks having different frequencies used and corresponding vibrating lengths of wire are noted as . by keeping the tension constant ( .

viii. It is observe that . constant, for constant value of tension and mass per unit length .

constant

i.e., , if and are constant.

Thus, the first law of a vibrating string is verified by using a sonometer.

6.9 Bents

What are beats?

Ans: One waxing and successive waning together constitute one beat.

Explain what is Doppler effect in sound and state its any ‘four’ applications.

Ans:

i. When a source of sound and the listener are in relative motion, the listener detects a sound whose frequency is different from the actual or original frequency of the sound source. This is Doppler effect.

ii Application of Doppler effect:

a. The principle of Doppler effect is used by traffic police to determine the speed of a vehicle to chock whether speed limit is exceeded.

b. SONAR works on the principle of Doppler effect for determining the speed of submarines using a sound source and sensitive microphones.

c. Doppler ultrasonography and echo cardiogram work on Doppler effect

d. Speed of an airplane can be determined using doppler RADAR.

Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.

Show that beats frequency is equal to frequency difference between two interfering waves.

Ans:

i. Consider two sound waves, having same amplitude and slightly different frequencies and .

ii Let be a point of the medium where they arrive in phase.

-in The displacement due to each wave at any instant of time at that point is given as

iv. Let us assume for simplicity that the listener is at .

and

v. According to the principle of superposition of wàves,

[By using formula,

Rearranging the above equation, we get

vi. Let

This is the equation of a progressive wave having frequency ‘ ‘ and amplitude ‘ ‘. The frequency ‘ ‘ is the mean of the frequencies and of arriving waves while the amplitude varies periodically with time.

vii. The intensity of sound is proportional to the square of the amplitude. Hence the resultant intensity will be maximum when the amplitude is maximum.

viii. For maximum amplitude (waxing),

i.e.,

Thus, the time interval between two successive maxima of sound is always .

Hence, the period of beats is .

The number of waxing heard per second is the reciprocal of period of waxing.

Beat frequency in waxing,

ix. The intensity of sound will be minimum when amplitude is zero (waning):

For minimum amplitude, ,

Therefore, time interval between two successive minima is also .

The number of waning heard per second is the reciprocal of period of waning.

Period of beat,

Beat frequency in waning,

x. Beats frequency is equal to frequency difference between two interfering waves.

Numericals

6.2 Progressive Wave

A simple harmonic progressive wave is given by the equation in S.I. units. Find the amplitude, frequency, wavelength and velocity of the wave. [Mar 10]

Solution:

Given: To find:

Formulae:

Calculation: Comparing with formula (i),

Ans: For the given wave amplitude, frequency, wavelength and velocity are respectively, , and .

The equation of simple harmonic progressive wave is given by , where all quantities are in . I. units. Calculate the displacement of a particle at from origin and at the instant 0.1 second.

Solution:

Given: ,

To find: displacement

Formula:
Calculation: Comparing with formula,

Ans: The displacement of the particle from origin is .

6.7 Harmonics and Overtones

The velocity of sound in air at room temperature is . An air column is in length. Find the frequency of third overtone in a pipe, when it is (i) closed at one end (ii) open at both ends.

Solution:

Given: ,

To find:

Frequency of third overtone .

Formulae: i. ii.

Calculation: From formula (i),

Fundamental frequency of a pipe closed at one end,

In this case, only the odd harmonics are present.

Using, we get,

From formula (ii),

Fundamental frequency of a pipe open at both ends is,

In this case, all harmonics are present.

Using, we get,

Ans: i. The frequency of third overtone in the pipe when it is closed at one end is .

ii. The frequency of third overtone in the pipe when it is open at both end is .

The consecutive harmonics of an air column closed at one end are and respectively. Find the fundamental frequency of the similar air column but open at both ends.

Solution:

Let fundamental frequency of closed pipe and frequency of and .

Since only odd harmonics are present,

and

Now, and

On solving,

The two given frequencies correspond to and overtones or and harmonics.

Now, for given air columns of same length but one closed at one end and the other open at both ends,

Ans: The fundamental frequency of the similar air column open at both ends is .

A pipe which is open at both ends is long and has an inner diameter . If the speed of sound in air is , calculate the fundamental frequency of air column in that pipe.

Solution:

Given: ,

To find: Fundamental frequency (n)

Formula:

Calculation: From formula,

Am: The fundamental frequency of air column is .

A stretched wire emits a fundamental note of frequency . Keeping the stretching force constant and reducing the length of wire by , the frequency becomes . Calculate the original length of wire.

Solution:

Given:

To find: Length

Formula:

Calculation: From first condition,

From second condition,

Dividing equation (1) by (2)

Ans: The original length of wire is .

The fundamental frequency of an air column in a pipe closed at one end is in unison with the third overtone of an open pipe. Calculate the

Solution:

ratio of lengths of their air columns. [Mar 16]

Given: Fundamental frequency for closed pipe Third overtone of open pipe.

To find: Ratio of lengths of air columns in both the pipes

Formulae: i. Fundamental frequency of pipe closed at one end:

ii. Fundamental frequency of pipe open at both ends:

iii. Third overtone of open pipe:

Calculation: Fundamental frequency of closed pipe is same as third overtone of open pipe.

Ans: The ratio of lengths of air columns in closed pipe and open pipe is 1:8.

A tube open at both ends has length . Calculate the fundamental frequency of air column. (Neglect end correction. Speed of sound in air is ).

Solution:

Given:

To find: Fundamental frequency of air column (n)

Formula:

Calculation: From formula,

Ans: The fundamental frequency of air column is 351.2 Hz.

A transverse wave is produced on a stretched string long and fixed at its ends. Find the speed of the transverse wave, when the string vibrates while emitting second overtone of frequency .

Solution:

Given:

To find: Speed (v)

Formula:

Calculation: In overtone, 3 loops are formed.

From formula,

Ans: The speed of the transverse wave is .

What should be tension applied to a wire of length and mass 10 gram, if it has to vibrate with fundamental frequency of ?

Solution:

Given:

To find: Tension (T)

Formula:

Calculation: Linear density of wire,

From formula,

Ans: The tension to be applied to the wire is .

The length of an air column for a fundamental mode in a resonance tube is and that for second resonance is . Find the end correction.

Solution:

Given: ,

To find: End correction (e)

Formula:

Calculation: From formula,

Ans: The end correction is or .

A wire length and mass is in unison with a tuning fork of frequency . Calculate the tension produced in the wire.

Solution:

Given: ,

Mass per unit length

To find: Tension

Formula:

Calculation: From formula,

Ans: Tension produced in the wire is .

A pipe open at both the ends has a fundamental frequency of . The first overtone of a pipe closed at one end has the same frequency as the first overtone of the open pipe. How long are the two pipes?

[Velocity of sound in air ]

Solution:

Given:

To find: Length of pipes and

Formulae: i.

ii.

Calculation: From formula (i),

From formula (ii),

Ans: i. The length of open pipe is .

ii. The length of one end closed pipe is .

6.8 Sonometer

A sonometer wire is in unison with a tuning fork when stretched by a weight of specific gravity ‘nine’. On completely increasing the weight in water, wire produces 4 beats per second with the fork. Calculate the frequency of the fork.

Solution:

Given: Specific gravity ,

Beat frequency

To find: Frequency

Formula:

Calculation: Let, weight of body in air, apparent weight of body in water, frequency of fork

Specific gravity,

From formula,

Fundamental frequency of vibration of wire,

and

Beat frequency

From (i) and (ii),

Ans: The frequency of the fork is .

A long sonometer wire vibrates with frequency of in fundamental mode, when it is under tension of . Calculate linear density of the material of wire.

Solution:

Given:

To find: Linear density

Formula:

Calculation: Using formula,

Ans: Linear density of material of wire is .

A sonometer wire is in unison with a tuning fork of frequency when it is stretched by a weight. When the weight is completely immersed in water, 8 beats are heard per second. Find the specific gravity of the material of the weight.

Solution:

Given: When wire is stretched by a weight hanging vertically, ,

Let frequency of wire when the weight is immersed in water producing 8 beats per second

To find: Specific gravity

Formula:

Calculation: From formula,

Ans: The specific gravity of material of the weight is 8.09 .

A stretched sonometer wire is in unison with a tuning fork. When the length of the wire is increased by , the number of beats heard per second is 10 . Find the frequency of the

Solution:

tuning fork.

[Mar 19]

Given:

To find:

Formula: For two wires of same material,

Calculation:

In first case,

In second case,

Since, , then

Given that,

Ans: The frequency of the fork is .

A sonometer wire 1 metre long weighing is in resonance with a tuning fork of frequency . Find tension in the sonometer wire.

Solution:

Given: ,

To find: Tension in the wire (T)

Formula:

Calculation: From formula,

Ans: The tension in the sonometer wire is .

6.9 Beats

Wavelength of two sound notes in air are and . Each of these notes produce 4 beats per second with a third note of fixed frequency. Find the frequency of third note and velocity of sound in air.

Solution:

Given:

To find: Frequency of third note (n),

Formula:

Calculation: For two notes in air,

From the given condition of the problem, and

From formula,

Ans: i. The frequency of the third note is .

ii. Velocity of sound in air is .

Wavelengths of two notes in the air are and . Each of these notes produces 8 beats per second with a tuning fork of fixed frequency. Find the velocity of sound in the air and frequency of the tuning fork.

Solution:

Given:

To find: Frequency (n),

Velocity of sound

Formula:

Calculation: Using formula,

For two notes in air,

Considering the given condition in problem,

and

Using formula,

Ans: i. The frequency of tuning fork is

ii. The velocity of sound in air is .

In a set, 21 tuning forks are arranged in a series of decreasing frequencies. Each tuning fork produces 4 beats per second with the preceding fork. If the first fork is an octave of the last fork, find the frequencies of the first and tenth fork.

Solution:

Given: beat

To find: i. Frequency of first fork

ii. Frequency of tenth fork

Formula:

Calculation:

i When tuning forks are arranged in the decreasing order of frequencies, the frequency of the tuning fork is,

As frequency of first fork is an octave of last,

From equation (1),

Ans: i. The frequency of the first fork is . ii. The frequency of the tenth fork is .

A set of 48 tuning forks is arranged in a series of descending frequencies such that each fork gives 4 beats per second with preceding one. The frequency of first fork is 1.5 times the frequency of the last fork, find the frequency of the first and tuning fork.

Solution:

Given:

To find: i. Frequency of first tuning fork

ii. Frequency of tuning fork

Formula:

Calculation:

i. When tuning forks are arranged in the decreasing order of frequencies, the frequency of the tuning fork is given by,

As frequency of first fork is 1.5 times of last,

From equation (1),

ii. For fork,

Ans: i. Frequency of first fork is .

ii. Frequency of fork is .

Two sound notes have wavelengths and in the air. These notes when sounded together produce 8 beats per second. Calculate the velocity of sound in the air and frequencies of the two notes.

Solution:

Given:

Beat frequency, beats per second

Std. XII Sci. Board Questions with Sollitions (Physics)

To find:	i.	Velocity ,
	ii.	Frequency

Formula:

Calculation: From formula,

Ans: i. The velocity of sound in air is .

ii. The frequencies of the two notes are and .

A set of 12 tuning forks is arranged in order of increasing frequencies. Each fork produces ‘ ‘ beats per second with the previous one. The last is an octave of the first. The fifth fork has a frequency of . Find ‘ ‘ and frequency of the first

Solution:

and the last tuning forks.

[July 17]

Given:

To find: Number of beats ,

Frequency of last fork

Formula:

Calculation: From formula,

Substituting in equation (1),

Ans: . The value of is 6 beat/s

ii. The frequency of the first and last tuning fork is and respectively.

The wavelengths of two sound waves in air are and . They produce 10 beats per second. Calculate the velocity of sound in air.

Solution:

Given: ,

Beat frequency, beats per second

Ans: The velocity of sound in air is .

Two tuning forks of frequencies and are sounded together to produce sound wave. The velocity of sound in air is . Calculate the difference in wavelengths of these waves.

Solution:

Given: ,

To find: Difference of wavelength of the waves

Calculation: From formula,

Ans: The difference in wavelength is .