## Rajasthan Board RBSE Class 11 Physics Chapter 9 Wave Motion

### RBSE Class 11 Physics Chapter 9 Textbook Exercises With Solutions

#### RBSE Class 11 Physics Chapter 9 Very Short Answer Type Questions

Question 1.

What is transferred in wave motion?

Answer:

In wave motion, energy is transferred through the motion of particles of the medium.

Question 2.

If the length of a stretched string is doubled and the tension is increased to four times then what would be the relation between new frequency and the old frequency?

Answer:

Question 3.

Write the relation between angular frequency, angular wave number and wave velocity.

Answer:

Question 4.

What does the number of vibrations executed by the particle of a medium in one second is called?

Answer:

Frequency.

Question 5.

What is the time to complete one the vibration called?

Answer:

Time period.

Question 6.

Write the equation of wave velocity.

Answer:

Wave velocity υ = nλ; where n = frequency, λ = wavelength.

Question 7.

What is the velocity of sound in air at standard temperature and pressure?

Answer:

Velocity of sound in air at standard temperature and pressure is 332 m s^{-1}.

Question 8.

What will be the change in the intensity of a wave if its amplitude is reduced to half?

Answer:

Thus the intensity is reduced to one fourth of its original value.

Question 9.

From which medium a wave is reflected so that the phase changes for a reflected wave?

Answer:

When a wave is reflected from a denser medium, its phase changes by π.

Question 10.

What will be the frequency of the beats if two tuning forks of frequency 400 and 402 are vibrated together?

Answer:

Difference is the frequency of the turning forks

n = n_{1} ~ n_{2} = 400 ~ 402

or n = 2 beats/second

Question 11.

What will be the ratio of fundamental frequencies of an open and closed organ pipes of same length?

Answer:

Fundamental frequency of an open pipe

Question 12.

In which closed or open organ pipe only odd harmonics are generated?

Answer:

In closed organ pipe.

Question 13.

What is the maximum displacement from mean position called?

Answer:

Amplitude.

Question 14.

Is there transfer of energy by the standing waves?

Answer:

No.

Question 15.

Which waves are produced in a resonance air column?

Answer:

Longitudinal standing waves.

Question 16.

How much is the distance between an antinode and its consecutive node?

Answer:

Distance between an antinode and its consecutive node = λ/4.

Question 17.

What is the effect of temperature on velocity of sound?

Answer:

υ ∝ \(\sqrt { T }\) i. e., velocity increases with increase in temperature. Formula for variation in temperature.

υ_{t} = υ_{0} + 0.61 t

Question 18.

From loud or shrill sound whose pitch is more?

Answer:

Pitch is more for shrill sound.

Question 19.

Can Doppler’s effect be seen in the whistle of an aeroplane moving with the ultrasonic velocity?

Answer:

No, because to realise Doppler’s effect the velocity of the sound source and listener should be less than the velocity of sound.

#### RBSE Class 11 Physics Chapter 9 Short Answer Type Questions

Question 1.

What are elastic waves?

Answer:

The motion in a medium in which, when particles are displaced, a force proportional to the displacement acts on the particles to restore them to their priginal position is called an elastic wave. If a material has the property of elasticity and the particles in the medium are set into vibratory motion, an elastic wave will be propagated. For example, a gas is an elastic medium and sound is transmitted through a gas as an elastic wave.

Question 2.

For effective propagation of wave what should be the properties of a medium?

Answer:

For the effective propagation of the waves the medium should have the following properties :

1. The medium should have the elasticity property so that if a particle is displaced then it comes back to the original (initial) position and displaces the particle close to it.

2. For efficient transfer of energy the material medium should have the property of inertia so that it can collect, the energy otherwise the energy would be lost.

3. Every medium resists the disturbance of the motion; which is called resistance. Due to this the energy of oscillation of the particles reduces. If this resistance is more then the dimension of the disturbance becomes zero which means the propagation of energy is for a very small distance. Hence, the resistance of the medium should be very small.

Question 3.

Define wave propagation constant.

Answer:

Wave Propagation Constant (k) : During vibration the phase difference between the particles which are situated at a unit distance is called the wave propagation constant. It is also called the angular wave number. Since, the phase difference is 2π in the particles situated at λ distance. Hence for unit distance;

Question 4.

Differentiate between transverse and longitudinal waves.

Answer:

Question 5.

Write about the reflection of waves.

Answer:

Reflection of Transverse Wave:

To understand the reflection of the transverse waves we study a stretched string as shown in the figure 9.9 whose one end is tied to a rigid base, and consider a pulse (disturbance) moving towards the stable end. This disturbance would put force on the rigid base and as a reaction the rigid base would put force on the string. Due to this the direction of the disturbance is changed which is called reflected disturbance. The direction of reflected disturbance is opposite to the direction of incident disturbance i.e., crest is reflected in the form of trough and trough is reflected in the form of crest but there is no change in their shape.

Reflection of Longitudinal Waves:

To understand the reflection of longitudinal waves we study about 4 balls of light weight (less density substance) of same mass and 4 balls of heavy weight (high density substance) and of the same mass as shown in the figure (9.13). These balls are placed on a horizontal surface which is frictionless and all balls are joined together by small and same pieces of spring. The first four balls from the left hand side are light and the remaining four are heavy.

Question 6.

Differentiate between progressive and standing waves with their definitions.

Answer:

Question 7.

What do you understand by superposition of waves? Explain.

Answer:

Superposition of Waves:

What would happen in a situation a progressive wave propagates in a medium and if we consider similarly many waves propagating in the same direction or opposite direction?

In this situation to study the displacement of the particle, we analyse two disturbances moving in the- opposite direction at the same time in a stretched string as shown in the figure. According to the situation (a); two disturbances which are in the same phase, are moving towards each other in opposite direction and at the place of their meeting the displacement of the particle is different than the initial displacement. Since the waves are in the same phase it increases and the disturbances moving ahead then take the original position. Similarly according to situation (b); if the disturbances are in opposite phase and are imposed; then the resultant displacement in the diagram is different from the initial displacement and with time as the disturbance moves ahead then displacement takes its original position.

Hence, we can see that when on a particle at the same time two or more than two waves are imposed, then they effect the displacement of the particle. Therefore; “when more than one wave is imposed on a particle at the same time, it is called superposition of waves.”

As a result of superposition of waves the particle instantaneously executes motion under the total resultant effect of the imposed waves and the resultant displacement is given by the principle of superposition,

Question 8.

Analyse the Laplace’s correction for the wave velocity expression in gas.

Answer:

Laplace Correction:

According to Laplace when sound waves travel in air then there is compression and rarefaction of air. In the place of compression the air particles are closer hence become hot and in the position of rarefaction the particles are further apart hence temperature reduces; and this process of compression and rarefaction is so continuous that heat is not able to move out from the medium and control its temperature. Hence, temperature is not constant in this process. From gas equation for isothermal changes.

Question 9.

Write the rules for the transverse vibrations in a stretched string?

Answer:

We know that vibrations in a stretched string is an example of transverse standing waves. To understand it we study a string which is tied between two stationary points and is stretched. If T is the tension in the string and mass of unit length is m, then the velocity of waves produced in the string from equation (9.5).

When this string is stretched in the perpendicular direction slightly and then left. Then the transverse progressive waves start moving. These waves are reflected from the rigid ends of the string. Therefore, by superposition of incident and reflected waves transverse standing waves are generated and these waves remain till their energy is not destroyed due to friction, etc. Since both the ends of the string are tied up hence they are always nodes.

When the string is stretched from middle and is left then the string vibrates in one part. Both the ends of the string are nodes and the middle point is antinode figure (9.15). This is the simple method of producing standing wave in a stretched string in which the string vibrates in a loop.

Question 10.

How are standing waves produced in sonometer? Clearly explain.

Answer:

When the string which is stretched between the two fixed points is plucked at its center, vibrations are produced and it moves out in opposite directions along the string. Because of this a transverse wave travels along the string and generates the standing waves.

Question 11.

Define damped vibrations and maintained vibrations.

Answer:

When an object which execute vibration motion is slightly displaced from its stable position and is left then due to the effect of restitution force it vibrates with its natural (fundamental) frequency. This type of vibration is called the “free vibration”. In ideal free vibration the amplitude of the object is constant state of. But in reality when the object excutes free vibrations then it experiences some frictional force in the medium in which it is vibrating. As a result some part of its energy is lost as heat. Hence, the amplitude of the object decreases with time. Those types of vibrations whose amplitude decreases with time are called the “damped vibrations.”

If somehow we can provide the same energy to the vibrating object as to which it has lost; then the object keeps on vibrating with the same (constant) amplitude. This type of vibration is called the “maintained vibrations”.

A vibrating object forces another object (which has the tendency to vibrate) to vibrate. It is also possible if their fundamental frequencies are different. This type of vibration is called the “forced vibrations”. This type of vibrations end very soon because in this situation the second vibrating object does not vibrate with its fundamental frequency but vibrates with the frequency of the first vibrating object.

When the frequency of first vibrating object is equal to the fundamental frequency of the second vibrating object then the amplitude of forced vibrations increases relatively. This is called the “resonance”. Every impulse of frequency force helps the system state to vibrate. Due to this the amplitude of the system keeps on increasing.

Question 12.

If the wavelength of a wave is 2500 A then what will be the wave number?

Answer:

Question 13.

Define antinodes and nodes.

Answer:

1. Some medium particles (A_{1},A_{2},A_{2},A_{4} and A_{5}) always vibrate with the maximum amplitude on both the sides of the mean position. They are called antinode. Velocity (dy/dt) of particles at antinodes is, maximum and pressure or change in density is zero.

[∵ \(\frac { dy }{ dx } \) = 0]

2. Particles (N_{1}, N_{2}, N_{3}, N_{4}) between the consecutive antinodes are always stationary. These are called node. These are positioned at same distance. Velocity of particles [dy/dt] at nodes is zero and the pressure or density change is maximum.

Question 14.

On what factors does the intensity of a wave depends upon?

Answer:

Intensity of a wave

I = 2π^{2} n^{2} α^{2} ρ.υ

∴ The intensity of wave depends on :

(i) Frequency; I ∝ n^{2}

(ii) Amplitude I ∝ α^{2}

(iii) Density of a medium 7 ∝ ρ

(iv) Velocity of a wave I ∝ υ

Question 15.

How can the frequency of a tuning fork be calculated by the methods of beats?

Answer:

In incidents related to sound, beats are very important and has many uses. For tuning of musical instruments this is used. If we wish to listen only one type of note from the musical instrument then it is only possible if it does not have beats. When beats are not present the musical instrument become sametone. With the help of this the frequency of unknown tuning fork is also known.

By the method of beats the calculation of the frequency of unknown tuning fork.

To calculate the frequency of unknown tuning fork, an approximately same but known frequency tuning fork is taken because to observe the beats of sound waves is necessary that the difference in the frequency of both the forks is not much otherwise beats will be formed so fast that to hear them will not be possible. Both the tuning forks are vibrated together and per second the number of beats is calculated. Suppose known frequency of the tuning fork is n and the obtained beats is ∆n then the frequency of unknown tuning would be n + ∆n or n – ∆n.

Question 16.

What are the limitations of Doppler’s Effect?

Answer:

Limitations of Doppler’s Effect:

To observe (view) Doppler’s effect it is necessary that the source, the observer and the medium the velocity of all should be less than the velocity of sound. If their velocity is more than the velocity of sound then the wave velocity graph is distorted and shocking waves are generated. Its example is Jet plane, when the speed of jet plane is more than the speed of sound then Doppler’s effect is not seen.

Question 17.

On what factors, does the Doppler’s Effect in sound waves depend upon?

Answer:

Doppler’s Effect in sound waves depends upon the following factors :

- Velocities of the source and the observer
- Relative velocities of the source and the observer.
- Whether the source or/and the observer is in motion.
- Velocity of the medium.

Question 18.

With what velocity does source move towards the observer such that the apparent frequency become double?

Answer:

When the source moves towards the stationary observer, the apparent frequency is.

∴ Velocity of source should be half the velocity of the sound.

Question 19.

How can we calculate the velocity of submarine by Doppler’s effect?

Answer:

The most important use of Doppler’s Effect in sound is to calculate the velocity of submarines. Sound wave are sent from the seashore towards the sea; they are reflected back from the enemy submarines and are gained back at sea-shore (SONAR station). Now by calculating the change in the wavelength of the reflected wave the velocity of submarine is calculated as follows.

There is no change in the wavelength of the waves reaching the submarine from the SONAR station because in this case source is stationary but there is change in the wavelength of the reflected waves because for the reflected waves submarine will act as sound source. If submarine (sound source) is coming towards the SONAR station then the apparent wavelfength will be:

#### RBSE Class 11 Physics Chapter 9 Long Answer Type Questions

Question 1.

What do you understand by wave motion? Clearly explain. Establish the expression and the one dimensional differential equation for a progressive wave.

Answer:

In the physical world it is compulsory for us to transfer energy from one place to another in many cases. “Wave motion is a technique which transfers energy from one point to another, often with no permanent displacement of the particles of the medium : that is, with little or no associated mass transport.” Almost all the branches of Physics have their relation with the wave motion. Waves can be broadly classified into three types :

1. Electromagnetic Waves : Those waves which do not require any medium to transport their energy from one place to another are called electromagnetic waves. Thermal radiation, light waves, X-rays, radio waves, gamma rays, etc. are examples of electromagnetic waves.

2. Matter Waves : When the matter particles like electron, proton, neutron, etc. are in motion, then the waves related to these moving particles are called matter waves. In electron microscope the matter waves associated with electron are used.

3. Mechanical Waves : Those waves which require the material medium for their propagation or transfer of energy are called mechanical waves. Sound waves are an example of mechanical waves.

In this chapter we will study about mechanical waves which will be called waves with reference to this chapter. There are mainly two methods for transfer of energy in Mechanics.

When in any medium the wave propagates then the medium particle vibrates and this vibrational motion is called simple harmonic motion. Medium particles vibrate in same or different phases.

Suppose a wave is moving as shown in diagram from point O to the right hand side.

Particles on the right hand side of O are disturbed after some time hence due to this their motion is lagging. Displacement equation for a particle at point P which is at a distance x from origin O is;

y = α sin (ω t – Φ)

Here, Φ is the phase difference between particles at O and P. Since, point P is at a distance x from O and the phase difference between particles at distance X is 2π.

Hence, phase difference between the particles at distance x is

Equations (9.9) and (9.10) represent progressive wave Equation in the positive x direction.

Similarly, progressive wave equation in negative x direction will be given by; y = α sin (ωt + kx).

Again from equation (9.9)

y = α sin (ωt – kx)

Differentiating the above equation with respect to time;

υ = \(\frac { dy }{ dt } \) = ω α cos (ω t – kx) …(9.11)

This is the expression for velocity of the particle. Hence, it is clear that the velocity of the particle is different from wave velocity. Velocity of the particle is dependent on time (t) and position (x) of the particle, whereas wave velocity is constant. From equation (9.11) the maximum value of the velocity of a particle is ω a.

Again differentiating equation (9.11) with respect to time;

Equation (9.12) shows that the acceleration of a particle is directly proportional to displacement and is in opposite direction. This is the necessary condition of simple harmonic motion. Differentiating ‘equation (9.9) twice relative to position (x)

Equation (9.14) is called one dimensional differential equation of the wave.

Question 2.

Derive the formula for velocity of wave in liquid.

Answer:

We have studied earlier that due to disturbance in stretched string transverse waves are generated. Hence, to calculate the velocity of transverse waves, let us assume a string, whose unit length mass is m, and tension is T. A disturbance is in motion from left towards right with velocity υ. We can also assume that the observer is moving in the direction of disturbance with the same velocity v, then the vibration will appear stable to the observer and the string will appear moving in the opposite direction.

Now analyse the small part δl; due to disturbance if there is small displacement in the string then this small part δl; can be considered as a part of the circle with radius R and the mass of this small part will be mδl.

On this small part as shown in the diagram, the component of tension towards the center is 2T sin θ which will provide the necessary centripetal force. Therefore;

This is the equation for velocity of a transverse wave which shows that the velocity of the wave in a string is dependent on the tension and the mass of unit length and is not dependent on amplitude of the wave and its wavelength. In the above situation string is considered to be ideal (that is totally elastic and same density) and while doing vibration motion there is no change in the length of the string.

Question 3.

Derive the formula for velocity of the transverse waves in a uniform stretched string.

Answer:

We have studied earlier that due to disturbance in stretched string transverse waves are generated. Hence, to calculate the velocity of transverse waves, let us assume a string, whose unit length mass is m, and tension is T. A disturbance is in motion from left towards right with velocity υ. We can also assume that the observer is moving in the direction of disturbance with the same velocity v, then the vibration will appear stable to the observer and the string will appear moving in the opposite direction.

Now analyse the small part δl; due to disturbance if there is small displacement in the string then this small part δl; can be considered as a part of the circle with radius R and the mass of this small part will be mδl.

On this small part as shown in the diagram, the component of tension towards the center is 2T sin θ which will provide the necessary centripetal force. Therefore;

This is the equation for velocity of a transverse wave which shows that the velocity of the wave in a string is dependent on the tension and the mass of unit length and is not dependent on amplitude of the wave and its wavelength. In the above situation string is considered to be ideal (that is totally elastic and same density) and while doing vibration motion there is no change in the length of the string.

Question 4.

Prove that in a string the ratio of velocity of longitudinal waves and transverse waves is

Answer:

To calculate the velocity and propagation of longitudinal waves we take a tube having a piston containing, compressed liquid, which can be divided into similar lines in many layers of same liquid. Where the lines are closer the pressure and density of the liquid is more and where the lines are farther apart the pressure and density of the liquid is less. Here we consider the liquid as a continuous medium and will ignore that in reality it is made up of atoms which are moving in different directions with different velocities.

If the piston is pushed inside the tube then the liquid will be compressed and it would thus increase the pressure and density. This layer of the compressed liquid will move ahead and would generate compression in another layer. If the piston is again pulled outwards then due to this the liquid will spread and its pressure and density will decrease which means there would be rarefaction. This way if piston is continuously pulled and pushed then the disturbance would move ahead in the tube in the form of rarefaction and compression. Hence, it behaves as a longitudinal wave.

Suppose the piston is pushed inside and a disturbance is generated which moves in the form of a compression and is moving towards the right with υ velocity. For simplicity, we assume that in this compression area, the pressure and density of the liquid is constant. Now if the observer is also considered to be moving in the direction of compression with same velocity then the liquid medium will be assumed to be moving in opposite direction to the compression with the υ velocity and for observer the compression would be at rest, In this situation when liquid with velocity υ moving towards the compression area collides with the compression area then the pressure at front end will be more than the pressure at the last end. Suppose that A P is the pressure difference between both the ends.

Due to this in this area B this element (liquid medium) would be compressed and in this area its velocity is some less υ – ∆υ and when this element will come out of compression area then it would again gain its volume. Due to this A P will move backwards and accelerated and due to this its velocity will again become υ. This is how this small part will reach position C.

When liquid element enters the compression area then the working resultant force on it (right side).

F = Pressure (P) × Area (A)

= (P + ∆P) A – PA

= ∆PA

where, A is the cross sectional area of the tube.

Out of compression area the length of this small part is υ∆t, where ∆t is the time taken by the small part to pass a point. Hence, volume of small part is υ∆t × A and mass p υA∆t where pis the: density of the liquid outside compression area. When this small part enters the Compression area then acceleration experienced by it is;

(∵ velocity reduces hence acceleration is negative).

Now by Newton’s second law of motion;

Force (F) = mass (m) × acceleration (α)

It is clear that velocity of wave- is dependent on property of the medium, elasticity constant and density.

Question 5.

Derive the relationship between the amplitude of a wave and its intensity.

Answer:

When in any medium a progressive wave propagates then it transmits the energy to the other particles of the medium gradually, due to which these particles successively start harmonic vibration. This energy is provided by wave source. This way energy is transferred from one part of the medium to another successively by progressive wave. Since, the medium particles execute simple harmonic motion hence the total energy of the particle is equal to the sum of its potential energy and kinetic energy. Therefore first we calculate the energy of unit volume i.e., energy density.

We know that the displacement equation of the particle moving with a velocity υ towards the x axis of a progressive wave is given as follows (see equation (9.10))

Question 6.

What are standing waves? Derive the resultant wave equation for standing waves.

Answer:

When two progressive waves having equal amplitude and frequency moving in opposite directions along the same line in a bounded medium and are superimposed then the wave produced as a result of, this superposition is not seen moving in any direction. These type of waves are called standing waves.

Similar to progressive waves standing waves are also of two types : Transverse and longitudinal. Transverse standing waves are generated in stretched strings and longitudinal standing waves are generated in air columns. The music from musical instruments is produced due to standing waves. Sitar, violin, Piano, Guitar, Iktara, etc. produce transverse standing waves and Flute, Whistle, Tabla, etc. produce longitudinal standing wave.

Question 7.

What is resonance? By drawing the diagram of resonance tube calculate the formula for velocity of sound in air.

Answer:

When an object which execute vibration motion is slightly displaced from its stable position and is left then due to the effect of restitution force it vibrates with its natural (fundamental) frequency. This type of vibration is called the “free vibration”. In ideal free vibration the amplitude of the object is constant state of. But in reality when the object excutes free vibrations then it experiences some frictional force in the medium in which it is vibrating. As a result some part of its energy is lost as heat. Hence, the amplitude of the object decreases with time. Those types of vibrations whose amplitude decreases with time are called the “damped vibrations.”

If somehow we can provide the same energy to the vibrating object as to which it has lost; then the object keeps on vibrating with the same (constant) amplitude. This type of vibration is called the “maintained vibrations”.

A vibrating object forces another object (which has the tendency to vibrate) to vibrate. It is also possible if their fundamental frequencies are different. This type of vibration is called the “forced vibrations”. This type of vibrations end very soon because in this situation the second vibrating object does not vibrate with its fundamental frequency but vibrates with the frequency of the first vibrating object.

When the frequency of first vibrating object is equal to the fundamental frequency of the second vibrating object then the amplitude of forced vibrations increases relatively. This is called the “resonance”. Every impulse of frequency force helps the system state to vibrate. Due to this the amplitude of the system keeps on increasing.

Resonance tube is that device with the help of which the state of resonance is produced by vibrating the air column and then calculating the velocity of sound in air.

As shown in figure 9.20 resonance tube has a one meter long glass tube T of 5 cm diameter. It is joined at the lower end by a rubber tube with the liquid (water) vessel (R). The vessel (R) is placed on a vertical stand on which it can be moved up and down and can be positioned. Tube T is filled with water so that normally its 2/3 part is filled with water. There is an air column in the tube T from the level of water to its mouth. Hence, this tube works as a closed organ pipe. Here, with the help of vessel R the length of air column can be changed. Therefore, its vibration frequency can be changed.

When a vibrating tuning fork is brought close to the open end of a tube T. Then forced vibrations are produced in the air column. Now if by changing the length of the air column, its fundamental frequency is kept same as tuning fork’s frequency, then the amplitude of vibration is maximum; which is called the state of resonance and hence high intensity sound (shrill) is heard. In this situation the length of the air column is calculated. This is called the resonant length. This process is repeated again and again and the minimum (smallest) resonant length is calculated. This is called the first resonant length, at this time the air column vibrates with its fundamental frequency. If first resonant length is l then according to the figure 9.21 (a).

Question 8.

What are beats? By mathematical analysis prove that the number of beats per second is equal to the difference of frequencies of the sources.

Answer:

If two waves are taken into consideration which are totally not similar but there is some difference in their frequency meaning the wavelength of both is not same but somewhat different. This type of two waves are if superimposed at some point then at some instant they can produce high sound being in same phase but at another instant there can be phase difference in them and there can be an instant at which at the same point they can be in opposite phase and produce zero intensity sound. After sometime they can be in the same phase and will produce high frequency sound. Similarly, at any point the phase difference will keep on changing and the intensity of sound will be high and low; would increase and decrease.

Figure (9.23) shows two waves superposition at the same place as per the time and formation of beats. For convenience the frequency of one wave is taken 5 and that of the other is taken 4. The wave with frequency 5 is shown by dotted line and wave with frequency 4 is shown by a straight line. Vertical axis shows displacement of particles from the mean position and horizontal axis shows time. Wave formed by the superposition of both is shown by thick black line. At instant A, both the waves are in the same phase and hence resultant displacement will be maximum and intensity of sound will be highest. After small time difference B, both the waves will be in opposite phase and hence will deduct the effect of each other and will make zero displacement and zero sound. Again at instant C, both the waves will be in same phase with maximum displacement and loud sound. Similarly, according to time at any place sound will be maximum or minimum. Hence, we can say as conclusion;

“Superposition of two slightly different frequencies waves produce sometimes high or low intensity sounds, this harmonic cycle is called beats.”

Phase difference of two high intensity sounds or two low intensity sounds is called time of beat, meaning in this time one beat is produced.

From the above figure one more conclusion is drawn that if first wave produces 5 vibrations in 1 second and second wave produces 4 vibrations in 1 second then in 1 second 1 beat is heard meaning frequency of beats is equal to the difference of the frequencies of two waves.

Question 9.

Describe Doppler’s effect for sound waves and calculate the formula for experienced frequency when :

(a) Source is moving towards the stationary observer.

(b) Observer is moving towards the stationary source.

Answer:

Therefore Doppler Effect is a phenomena, it is not only applicable on sound waves but also works for electro-magnetic waves. Here we will only study sound waves and the changes in frequency at different situations of the source and observer.

Suppose sound source is S and observer is O and source S is vibrating with a frequency n and υ is the velocity of sound in stable medium then:

The most important use of Doppler’s Effect in sound is to calculate the velocity of submarines. Sound wave are sent from the seashore towards the sea; they are reflected back from the enemy submarines and are gained back at sea-shore (SONAR station). Now by calculating the change in the wavelength of the reflected wave the velocity of submarine is calculated as follows.

There is no change in the wavelength of the waves reaching the submarine from the SONAR station because in this case source is stationary but there is change in the wavelength of the reflected waves because for the reflected waves submarine will act as sound source. If submarine (sound source) is coming towards the SONAR station then the apparent wavelfength will be:

Where υs is the velocity of source means that of submarine and λ is the velocity of sound in water and X is the original wavelength of the sound waves. Negative sign means submarine is coming closer and the wavelength of the reflected waves from it decreases. If submarine is going away from SONAR station then there would be positive sign. Hence, simple, formula for change in wavelength will be;

By using this formula, the velocity of submarine (υ_{s}) is calculated and the position of the submarine relative to the SONAR station is also calculated.

#### RBSE Class 11 Physics Chapter 9 Numerical Questions

Question 1.

Write down the one-dimensional differential equation of a wave and examine which of the following are possible equations of the one-dimensional wave :

(i) y = 2sinx cos υt

(ii) y = 5sin2x cos υt?

Solution:

This is a one-dimensional differential equation. Hence this solution is possible.

Question 2.

A source of sound of frequency 500 Hz is producing longitudinal waves in air. The distance between the two successive rarefactions is 0.64 m and the amplitude of the air particles vibrating is’0.002 m. Calculate the displacement of a particle at 10 m distance from the origin at t = 2 s in the direction of the wave.

Solution:

Given; n = 500Hz; λ = 0.64 m; α = 0.002 m

∴ υ = nλ = 500 × 0.64 = 5 × 64 = 320 m s^{-1}

Hence, the displacement equation of the longitudinal wave,

Question 3.

The equation of a wave propagating on a string is as follows :

y = 10 sin π(0.01x – 2.00t) where y and x are in cm and t is in second. Calculate the amplitude, frequency and velocity of the wave. Calculate the phase difference between the two particles at a distance 40.0 cm at any instant.

Solution:

Question 4.

The fundamental frequency of 1.0 meter long stretched steel wire is 250 Hz. Density of Steel is 8000 kg/m3.

(i) Calculate the speed of a transverse wave in the wire.

(ii) Calculate the longitudinal stress of the wire.

(iii) If the tension of the wire is increased by 2% then calculate the percentage change in the frequency.

Solution:

Question 5.

Calculate the wavelength of a longitudinal wave vibrating with 400 vibrations/second in a metal of density 5.5 × 10^{3} kg/m^{3}. The Young’s modulus of metal is Y = 8.8 × 10^{10}N/m^{2}.

Solution:

Given; ρ = 5.5 × 10^{3} kg m^{-3}; n = 400 Hz; λ = ?

y = 8.8 × 10^{10} N m^{-2}

Velocity of the wave in the metal

Question 6.

Calculate the wavelength, wave number and frequency of a wave whose propagation constant is 2.8 × 10^{4} per metre and its velocity is 400 m/s.

Solution:

Question 7.

Calculate the frequency, wave number and propagation constant of a light wave of wavelength 5000 Å.

Solution:

Question 8.

A simple harmonic wave is moving towards the positive ac-direction with 100 m/s velocity. Amplitude of wave is 22 cm and the frequency is 100 vibrations/second (Hz). Calculate the displacement, velocity and acceleration of a particle which is at a distance x = 2 m from the origin at t = 5 seconds.

Solution:

Given υ = 100 ms^{-1}; α = 22 cm = 0.22m; n = 100Hz

∴ Equation of simple harmonic wave moving in positive x- direction

Question 9.

Calculate the frequency of the fundamental tone of a 50 m long wire stretched by a weight of 10 kg where weight of 1 m long wire is 2.45 g.

( g= 980 cm/s^{2})

solution:

Question 10.

Calculate the frequency of the fundamental tone of a 100 cm long wire with 1.8 mm diameter (density 8.4) and stretched by 20 kg weight.

solution:

l = 100cm = 1.00m; D = 1.8mm;

∴ r = \(\frac { 1.8 }{ 2 } \) = 0.9mm = 0.9 × 10^{-3} m = 9 × 10^{-4} m

P = 8.4 × 10^{3} kg m^{-3};

The fundamental tone frequency = ?

Mass of the stretched wire,

m = volume × density = πr^{2}.lp

= 3.14 × (9 × 10^{-4} m)^{2} × 8.4 × 10^{3}

= 3.14 × 81 × 10^{-8} × 8.4 × 10^{3}

= 2136.456 × 10^{-5}

= 2.1385 × 10^{-2}

T = Mg = 20 × 9.8 = 2 × 98 N

∴ Fundamental frequency of the wire

Question 11.

A 25 cm stretched wire and a tuning fork has same frequency. If the length of wire is changed to 25.5 cm and the tension is same and 3 beats per second are produced. Calculate the frequency of tuning fork.

Solution:

Question 12.

A sonometer wire gives sound of frequency 150 Hz. If wire’s tension is changed in the ratio 9 : 16 and length in the ratio 1 : 2. Calculate the frequency of the new wire.

Solution:

Question 13.

Equation of transverse vibrations of a string tied at both the ends is

Where y and x are in m and t in s. Length of the wire is 1.5 m and mass 0.002 kg, then,

(i) Calculate the maximum displacement at x = 0.5 m.

(ii) Mark the places of nodes on the wire.

(iii) Calculate the wave velocity.

(iv) Calculate the velocity of the particle at x = 0.75 m and t = 0.25 s.

Solution:

Question 14.

41 tuning forks are so arranged that every tuning fork generates 5 beat/s to its nearest tuning fork. The frequency of the last tuning fork is double the frequency of first tuning fork. Calculate the frequency of first and last tuning fork.

Solution:

Let the frequency of first fork = x

As each fork generates 5 beats/s to its nearest fork, thus it will be in A.P.

Thus for u/st fork = [α + (n – 1) × 5]

Given n = 41, d 5

⇒ = [x + (41 – 1) × 5]

= x + 200

As per question,

frequency of last fork = 2 × frequency of first fork

x + 200 = 2x

2x – x = 200

x = 100 Hz

∴ Frequency of first fork = 100 Hz frequency of last fork = 2 × 100 = 200 Hz

Question 15.

Calculate the velocity of sound in that gas in which two waves of wavelengths 1 m and 1.01 m produce 10 beats in 3 s.

Solution:

Given, λ_{1} = 1.00 m; λ_{2} = 1.01 m

Frequency of beats n = \(\frac { 10 }{ 3 } \) beats/second

Question 16.

Calculate the frequency of the tuning fork which when vibrated with the tuning fork of frequency 256 Hz produces 6 beat/s and when vibrated with tuning fork having frequency 253 Hz produces 3 beats/s.

Solution:

Let A and B be two tuning forks, then frequency of A

n_{A} = 256 Hz

Tuning fork b produces 6 beats with fork B.

∴ n_{B} = n_{A} ± n = 256 ± 6 = 250 or 262 Hz

∵Fork B produces 3 beat/s with the other fork (frequency 253 Hz), it is only possible when.

n_{3} = 250 Hz

Question 17.

A tuning fork produces 4 beats/second from the stretched string of a sonometer of length 0.49 m and 0.50 m. Calculate the frequency of the tuning fork.

Solution:

Frequency of the sonometer wire ∝ \(\frac { 1 }{ l } \)

Frequency of the tuning fork = n

Then frequency of 0.49 m length of wire = n + 4

and frequency of 0.50 m length of wire = n – 4

Question 18.

Two engines cross each other in opposite direction. One engine produce 540 Hz sound. People sitting in the second engine will hear sound of which frequency before and after passing from one another? The velocity of both the engines is 40 m/s. Velocity of sound is 340 m/s.

Solution:

Question 19.

A train is moving with a velocity of 60 km/h towards a siren whose frequency of sound is 400 vibrations/s. People sitting in the train will hear which frequency sound? (Velocity of sound in air is 340 m/s).

Solution:

Question 20.

An engine’s whistle frequency is experienced as 5/6th part when it crosses a stationary observer. If velocity of sound in air is 330 m/s then calculate the velocity of engine.

Solution: