Maharashtra Board Class 11 Physics Chapter 8 Sound Solutions

1. Choose The Correct Alternatives

Question 1. A sound carried by air from a sitar to a listener is a wave of following type.
(A) Longitudinal stationary
(B)Transverse progressive
(C) Transverse stationary
(D) Longitudinal progressive
Answer: (D) Longitudinal progressive
In simple words: Sound waves travelling through air are longitudinal because the particles vibrate parallel to the wave's direction, and they are progressive because they carry energy forward.

🎯 Exam Tip: Understanding the fundamental nature of different wave types (longitudinal vs. transverse, stationary vs. progressive) is crucial for conceptual questions.

Question 2. When sound waves travel from air to water, which of these remains constant ?
(A) Velocity
(B) Frequency
(C) Wavelength
(D) All of above
Answer: (B) Frequency
In simple words: When a sound wave passes from one medium to another, its speed and wavelength change, but its frequency remains constant as it's determined by the source.

🎯 Exam Tip: Remember that frequency is a source characteristic and doesn't change when a wave enters a new medium, while velocity and wavelength do.

Question 3. The Laplace's correction in the expression for velocity of sound given by Newton is needed because sound waves
(A) are longitudinal
(B) propagate isothermally
(C) propagate adiabatically
(D) are of long wavelength
Answer: (C) propagate adiabatically
In simple words: Laplace corrected Newton's formula because sound propagation in air is a very fast process, making it adiabatic (no heat exchange) rather than isothermal (constant temperature).

🎯 Exam Tip: Know the key difference between isothermal and adiabatic processes and why Laplace's correction is significant for sound velocity in gases.

Question 4. Speed of sound is maximum in
(A) air
(B) water
(C) vacuum
(D) solid
Answer: (D) solid
In simple words: Sound travels fastest in solids because their particles are closely packed and strongly linked, allowing vibrations to transmit quickly and efficiently.

🎯 Exam Tip: Recall the order of sound speed: solids > liquids > gases, and understand the reasoning based on molecular spacing and intermolecular forces.

Question 5. The walls of the hall built for music concerns should
(A) amplify sound
(B) Reflect sound
(C) transmit sound
(D) Absorb sound
Answer: (D) Absorb sound
In simple words: Music halls are designed with sound-absorbing walls to prevent echoes and reverberation, ensuring that the original sound is clear and distinct for the audience.

🎯 Exam Tip: This question tests knowledge of acoustics and architectural design principles for sound quality. Absorbing sound reduces unwanted reflections.

2. Answer Briefly.

Question 1. Wave motion is doubly periodic. Explain.
Answer:
i. A wave particle repeats its motion after a definite interval of time at every location, making it periodic in time.
ii. Similarly, at any given instant, the form of a wave repeats itself at equal distances making it periodic in space.
iii. Thus, wave motion is a doubly periodic phenomenon, i.e., periodic in time as well as periodic in space.
In simple words: Wave motion is "doubly periodic" because it repeats itself in time (oscillation at a point over time) and in space (wave shape repeating over distance at an instant).

🎯 Exam Tip: Define periodicity in terms of both time (period, frequency) and space (wavelength) to fully explain this concept.

Question 2. What is Doppler effect?
Answer: The apparent change in the frequency of sound heard by a listener, due to relative motion between the source of sound and the listener is called Doppler effect in sound.
In simple words: The Doppler effect is the observed change in a sound's pitch (frequency) when the source of the sound, the listener, or both are moving relative to each other.

🎯 Exam Tip: Focus on the *relative motion* between source and listener as the cause of the apparent frequency change.

Question 3. Describe a transverse wave.
Answer: Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave is called transverse wave. Example: Ripples on the surface of water, light waves.
Characteristics of transverse waves:
1. All the particles of medium in the path of wave vibrate in a direction perpendicular to the direction of propagation of wave with same period and amplitude.
2. When transverse wave passes through the medium, the medium is divided into alternate crests i.e., regions of positive displacements and troughs i.e., regions of negative displacement, that are periodic in time.
3. A crest and an adjacent trough form one cycle of a transverse wave. The distance between any two successive crests or troughs is called wavelength 'λ' of the wave.
4. Crests and troughs advance in the medium and are responsible for transfer of energy.
5. Transverse waves can travel only through solids and not through liquids and gases. Electromagnetic waves are transverse waves, but they do not require material medium for propagation.
6. When transverse waves advance through a medium, there is no change of pressure and density at any point of the medium, but the shape changes periodically.
7. Transverse wave can be polarised.
8. Medium conveying a transverse wave must possess elasticity of shape, i.e., modulus of rigidity.
In simple words: A transverse wave causes medium particles to vibrate perpendicular to the wave's direction of travel, creating crests and troughs, like ripples on water.

🎯 Exam Tip: Clearly state the direction of particle vibration relative to wave propagation and mention examples like light waves or water ripples.

Question 4. Define a longitudinal wave.
Answer: A wave in which particles of medium vibrate in a direction parallel to the direction of propagation of the wave is called longitudinal wave. Example: Sound waves.
In simple words: A longitudinal wave causes medium particles to vibrate parallel to the wave's direction of travel, creating compressions and rarefactions, like sound waves in air.

🎯 Exam Tip: Emphasize the parallel vibration and relate it to the formation of compressions and rarefactions in the medium.

Question 5. State Newton's formula for velocity of sound.
Answer: Newton's formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.
ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
\(v = \sqrt{\frac{E}{\rho}}\) ....(1)
Where, E is the modulus of elasticity of medium and p is density of medium.
Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.
2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.
Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
\(V = \sqrt{\frac{P}{\rho}}\) (:: E = P)
This is the Newton's formula for velocity of sound in air.
2. But atmospheric pressure is given by,
P = hdg

\(\implies v = \sqrt{\frac{hdg}{\rho}}\) ....(2)
3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air p = 1.293 kg/m³
4. From equation (2) we have velocity of sound,
\[V = \sqrt{\frac{0.76 \times 13600 \times 9.8}{1.293}} = 279.9 \text{ m/s at N.T.P}\]
In simple words: Newton's formula calculates the velocity of sound based on the medium's elasticity and density, assuming an isothermal process for propagation in air.

🎯 Exam Tip: Memorize the formula \(v = \sqrt{E/\rho}\) and the key assumptions (isothermal propagation) made by Newton.

Question 6. What is the effect of pressure on velocity of sound?
Answer: Effect of pressure:
i. Let v be the velocity of sound in air when the pressure is P and density is p.
ii. Using Laplace's formula, we can write,
\(V = \sqrt{\frac{\gamma P}{\rho}}\) ....(1)
iii. If V be the volume of a gas having mass M then, \(\rho = \frac{M}{V}\)
iv. Substituting \(\rho\) in equation (1), we get,
\(V = \sqrt{\frac{\gamma PV}{M}}\) ....(2)
v. But according to Boyle's law,
PV = constant (at constant temperature)
Also, M and \(\gamma\) are constant.

\(\implies v = \text{constant}\)
vi. Hence, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.
vii. For a gaseous medium, PV= nRT.
Substituting in equation (2), we get,
\(V = \sqrt{\frac{\gamma nRT}{M}}\)
viii. Thus, even for a gaseous medium obeying ideal gas equation, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.
In simple words: Pressure does not affect the velocity of sound in a gas as long as the temperature remains constant, because changes in pressure are compensated by changes in density, keeping the ratio constant.

🎯 Exam Tip: Highlight that for an ideal gas, if temperature is constant, pressure changes are offset by density changes, maintaining constant sound velocity according to Laplace's corrected formula.

Question 7. What is the effect of humidity of air on velocity of sound?
Answer: Effect of humidity:
i. Let \(v_m\) and \(v_d\) be the velocities of sound in moist air and dry air respectively.

\(\implies v_m = \sqrt{\frac{\gamma P}{\rho_m}}\) and
\(v_d = \sqrt{\frac{\gamma P}{\rho_d}}\)
\[ \implies \frac{v_m}{v_d} = \sqrt{\frac{\gamma P}{\rho_m}} \times \sqrt{\frac{\rho_d}{\gamma P}} = \sqrt{\frac{\rho_d}{\rho_m}} \]
ii. Humid air contains a large proportion of water vapour. Density of water vapour at 0 °C is 0.81 kg/m³ while that of dry air at 0°C is 1.29 kg/m³. So, the density \(\rho_m\) of moist air is less than the density \(\rho_d\) of dry air i.e., \(\rho_m < \rho_d\).
iii. Thus \(\frac{v_m}{v_d} > 1\)

\(\implies v_m > v_d\)
iv. Hence, sound travels faster in moist air than in dry air. It means that velocity of sound increases with increase in moistness (humidity) of air.
In simple words: Sound travels faster in humid air than in dry air because water vapor is less dense than dry air, reducing the overall density of the medium and increasing sound velocity.

🎯 Exam Tip: Explain that humidity makes the air less dense, and since sound velocity is inversely proportional to the square root of density, increased humidity leads to increased speed.

Question 8. What do you mean by an echo?
Answer: An echo is the repetition of the original sound because of reflection from some rigid surface at a distance from the source of sound.
In simple words: An echo is a reflected sound that reaches the listener after a distinct delay, allowing it to be heard as a separate sound from the original.

🎯 Exam Tip: Clearly state that an echo is a *reflected* sound and requires a *sufficient distance* from a rigid surface for distinct perception.

Question 9. State any two applications of acoustics.
Answer: Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.
ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.
Medical applications of acoustics:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.
ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.
iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.
Underwater applications of acoustics:
i. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
ii. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
iii. Motion and position of submerged objects like submarine can be measured with the help of this system.
Applications of acoustics in environmental and geological studies:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor's wheels and supporting structures.
Reflected and refracted elastic waves passing through the Earth's interior can be measured by applying the principles of acoustics. This is useful in studying the properties of the Earth.
Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.
In simple words: Acoustics is the study of sound, and its applications include bats using echolocation, dolphins navigating underwater, medical imaging (ultrasound), and designing sound-efficient buildings.

🎯 Exam Tip: Provide diverse examples from nature, medicine, and engineering to illustrate the broad applications of acoustic principles.

Question 10. Define amplitude and wavelength of a wave.
Answer:
i. Amplitude (A): The largest displacement of a particle of a medium through which the wave is propagating, from its rest position, is called amplitude of that wave. SI unit: (m)
ii. Wavelength (\(\lambda\)): The distance between two successive particles which are in the same state of vibration is called wavelength of the wave. SI unit: (m)
In simple words: Amplitude is the maximum displacement of a medium's particle from its rest position, while wavelength is the distance between two consecutive identical points on a wave, both being key measures of wave characteristics.

🎯 Exam Tip: Define both terms precisely with their SI units and visualize them in a wave diagram.

Question 11. Draw a wave and indicate points which are (i) in phase (ii) out of phase (iii) have a phase difference of \(\pi/2\).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक तरंग को दर्शाता है जिसमें x-अक्ष पर विस्थापन और y-अक्ष पर दूरी दिखाई गई है। तरंग एक आवधिक गति दिखाती है जिसमें कई बिंदु जैसे A, B, C, D, E, F, G, H, I, J और X-अक्ष पर दर्शाए गए हैं जो तरंग के विभिन्न चरणों का प्रतिनिधित्व करते हैं।
i. In phase point: A and F; B and H; C and I; D and J
ii. Out of phase points: A and B, B and D, FI and J, E and F,
iii. Point having phase difference of \(\pi/2\): A and B; B and C; D; D and F; F and H; H and I; J and I
In simple words: Points on a wave are "in phase" if they have the same displacement and velocity direction, "out of phase" if their conditions are opposite, and have a "phase difference of \(\pi/2\)" if one is at a maximum while the other is at zero displacement.

🎯 Exam Tip: Use the provided diagram points to illustrate phase relationships; understanding the concepts of displacement, velocity, and oscillation number is crucial.

Question 12. Define the relation between velocity, wavelength and frequency of wave.
Answer:
i. A wave covers a distance equal to the wavelength (\(\lambda\)) during one period (T). Therefore, the magnitude of the velocity (v) is given by,
Magnitude of velocity = \(\frac{\text{Distance covered}}{\text{Corresponding time}}\)
ii. \(v = \frac{\lambda}{T}\) i.e., \(v = \lambda \times (\frac{1}{T})\) ............ (1)
iii. But reciprocal of the period is equal to the frequency (n) of the waves.

\(\implies \frac{1}{T} = n\) ............ (2)
iv. From equations (1) and (2), we get
\(v = n\lambda\)
i.e., wave velocity = frequency \(\times\) wavelength.
In simple words: The relationship between velocity (v), wavelength (\(\lambda\)), and frequency (n) of a wave is given by \(v = n\lambda\), meaning velocity equals frequency multiplied by wavelength.

🎯 Exam Tip: Memorize the fundamental wave equation \(v = f\lambda\) (or \(v = n\lambda\)) and be able to derive it from basic definitions of period and wavelength.

Question 13. State and explain principle of superposition of waves.
Answer: Principle: As waves don't repulse each other, they overlap in the same region of the space without affecting each other. When two waves overlap, their displacements add vectorially.
Explanation:
i. Consider two waves travelling through a medium arriving at a point simultaneously.
ii. Let each wave produce its own displacement at that point independent of the others. This displacement can be given as,
\(y_1\) = displacement due to first wave.
\(y_2\) = displacement due to second wave.
iii. Then according to superposition of waves, the resultant displacement at that point is equal to the vector sum of the displacements due to all the waves.

\(\implies \vec{Y} = \vec{Y_1} + \vec{Y_2}\)
In simple words: The principle of superposition states that when multiple waves meet at a point, the resultant displacement is the vector sum of individual displacements, as waves pass through each other without interference.

🎯 Exam Tip: Emphasize that waves add vectorially and pass through each other unaltered, making it a linear superposition.

Question 14. State the expression for apparent frequency when source of sound and listener are
i) moving towards each other
ii) moving away from each other
Answer:
i. Let,
n = actual frequency of the source.
\(n_0\) = apparent frequency of the source,
v = velocity of sound in air.
\(v_s\) = velocity of the source.
\(v_L\) = velocity of the listener.
ii. Apparent frequency heard by the listener is given by,
\(n = n_0 \left(\frac{v \pm v_L}{v \mp v_s}\right)\)
Where upper signs (+ ve in numerator and -ve in denominator) indicate that source and observer move towards each other. Lower signs (-ve in numerator and +ve in denominator) indicate that source and listener move away from each other.
iii. If source and listener are moving towards each other, then apparent frequency is given by,
\(n = n_0 \left(\frac{v+v_L}{v-v_s}\right)\) i.e., apparent frequency increases.
iv. If source and listener are moving away from each other, then apparent frequency is given by,
\(n = n_0 \left(\frac{v-v_L}{v+v_s}\right)\) i.e., apparent frequency decreases.
In simple words: The expressions for apparent frequency in the Doppler effect depend on the relative motion: when source and listener move towards each other, frequency increases, and when they move away, it decreases.

🎯 Exam Tip: Understand the sign conventions for the Doppler effect formula; moving towards implies an increase in frequency (numerator `+vL`, denominator `-vS`), and moving away implies a decrease (`-vL`, `+vS`).

Question 15. State the expression for apparent frequency when source is stationary and listener is
1) moving towards the source
2) moving away from the source
Answer: Let,
n = actual frequency of the source.
\(n_0\) = apparent frequency of the source,
v = velocity of sound in air.
\(v_s\) = velocity of the source.
\(v_L\) = velocity of the listener.
i. If listener is moving towards source then apparent frequency is given by,
\(n = n_0 \left(\frac{v+v_L}{v}\right)\) i.e., apparent frequency increases.
ii. If listener is receding away from source then apparent frequency is given by,
\(n = n_0 \left(\frac{v-v_L}{v}\right)\) i.e., apparent frequency decreases.
In simple words: When the source is stationary, the apparent frequency heard by a listener increases if the listener moves towards the source and decreases if the listener moves away from it.

🎯 Exam Tip: For a stationary source, only the listener's velocity affects the numerator of the Doppler effect formula, causing an increase or decrease in apparent frequency.

Question 16. State the expression for apparent frequency when listener is stationary and source is.
(i) moving towards the listener
(ii) moving away from the listener
Answer: Let,
n = actual frequency of the source.
\(n_0\) = apparent frequency of the source,
v = velocity of sound in air.
\(v_s\) = velocity of the source.
\(v_L\) = velocity of the listener.
i. If source is moving towards observer then apparent frequency is given by,
\(n = n_0 \left(\frac{v}{v-v_s}\right)\) i.e., apparent frequency increases.
ii. If source is receding away from observer then apparent frequency is given by,
\(n = n_0 \left(\frac{v}{v+v_s}\right)\) i.e., apparent frequency decreases.
In simple words: When the listener is stationary, the apparent frequency increases if the source moves towards the listener and decreases if the source moves away.

🎯 Exam Tip: For a stationary listener, only the source's velocity affects the denominator of the Doppler effect formula, leading to an inverse relationship with apparent frequency.

Question 17. Explain what is meant by phase of a wave.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र x-अक्ष पर दूरी के फलन के रूप में विस्थापन को दर्शाता है, जिसमें एक तरंग की आवधिक गति को विभिन्न बिंदुओं जैसे m, n, p, q, r, s, t, u, v, w और X द्वारा चिह्नित किया गया है। यह तरंग के विभिन्न चरणों और कणों के विस्थापन व वेग की दिशाओं को समझने में सहायक है।
i. The state of oscillation of a particle is called the phase of the particle.
ii. The displacement, direction of velocity and oscillation number of the particle describe the phase of the particle at a place.
iii. Particles r and t (q and u or v and s) have same displacements but the directions of their velocities are opposite.
iv. Particles having same magnitude of displacements and same direction of velocity are said to be in phase during their respective oscillations. Example: particles v and p.
v. Separation between two particles which are in phase is wavelength (\(\lambda\)).
vi. The two successive particles differ by '1' in their oscillation number i.e., if particle v is at its \(n^{th}\) oscillation, particle p will be at its \((n+1)^{th}\) oscillation as the wave is travelling along + X direction.
vii. In the given graph, if the disturbance (energy) has just reached the particle w, the phase angle corresponding to particle is 0°. At this instant, particle v has completed quarter oscillation and reached its positive maximum (sin \(\theta\) = +1). The phase angle \(\theta\) of this particle v is \(\frac{\pi^c}{2}\) = 90° at this instant.
viii. Phase angles of particles u and q are \(\pi^c\) (180°) and 2rcc (360°) respectively.
ix. Particle p has completed one oscillation and is at its positive maximum during its second oscillation.

\(\implies \text{phase angle} = 2\pi^c + \frac{\pi^c}{2}\)

\( = \frac{5\pi^c}{2}\)
x. v and p are the successive particles in the same state (same displacement and same direction of velocity) during their respective oscillations. Phase angle between these two differs by \(2\pi^c\).
In simple words: The phase of a wave particle describes its state of oscillation, including its displacement, direction of velocity, and how many oscillations it has completed at a given moment.

🎯 Exam Tip: Understand that phase represents the specific point a particle is at within its cycle, and identical phases are separated by a full wavelength or period.

Question 18. Define progressive wave. State any four properties.
Answer:
i. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
Properties of progressive waves are:
Amplitude, wavelength, period, double periodicity, frequency and velocity.
In simple words: A progressive wave is a disturbance that travels through a medium, continuously transferring energy from one point to distant points without the medium itself being permanently displaced.

🎯 Exam Tip: Focus on the energy transfer and the continuous propagation of the disturbance as defining characteristics of progressive waves.

Question 19. Distinguish between traverse waves and longitudinal waves.
Answer:

In simple words: Longitudinal waves cause particles to vibrate parallel to wave direction, creating compressions and rarefactions, while transverse waves cause particles to vibrate perpendicular, forming crests and troughs.

🎯 Exam Tip: Clearly differentiate between the direction of particle vibration and wave propagation, and remember common examples for each wave type.

Question 20. Explain Newtons formula for velocity of sound. What is its limitation?
Answer: Newton's formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.
ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
\(V = \sqrt{\frac{E}{\rho}}\) ....(1)
Where, E is the modulus of elasticity of medium and p is density of medium.
Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.
2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.
Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
\(V = \sqrt{\frac{P}{\rho}}\) (:: E = P)
This is the Newton's formula for velocity of sound in air.
2. But atmospheric pressure is given by,
P = hdg

\(\implies v = \sqrt{\frac{hdg}{\rho}}\) ....(2)
3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air p = 1.293 kg/m³
4. From equation (2) we have velocity of sound,
\[V = \sqrt{\frac{0.76 \times 13600 \times 9.8}{1.293}} = 279.9 \text{ m/s at N.T.P}\]
Limitations:
1. Experimentally, it is found that the velocity of sound in air at N. T. P is 332 m/s. Thus, there is considerable difference between the value predicted by Newton's formula and the experimental value.
2. Experimental value is 16% greater than the value given by the formula. Newton failed to provide a satisfactory explanation for the difference.
In simple words: Newton's formula for sound velocity considers an isothermal process but has a limitation because sound propagation is actually an adiabatic process, leading to a calculated value lower than the experimentally observed speed.

🎯 Exam Tip: Remember that the main limitation of Newton's formula is its incorrect assumption of isothermal conditions; Laplace's correction addresses this by using adiabatic conditions.

Question 20. Explain Newtons formula for velocity of sound. What is its limitation?
Answer:
Newton's formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.
ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \( \sqrt{\frac{E}{\rho}} \) ....(1)
Where, E is the modulus of elasticity of medium and p is density of medium.
Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.
2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.
Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \( \sqrt{\frac{P}{\rho}} \) ....(:: E = P)
This is the Newton's formula for velocity of sound in air.
2. But atmospheric pressure is given by,
P = hdg
∴ v = \( \sqrt{\frac{hdg}{\rho}} \) ....(2)
3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air p = 1.293 kg/m³
4. From equation (2) we have velocity of sound,
v = \( \sqrt{\frac{0.76 \times 13600 \times 9.8}{1.293}} \) = 279.9 m/s at N.T.P
Limitations:
1. Experimentally, it is found that the velocity of sound in air at N. T. P is 332 m/s. Thus, there is considerable difference between the value predicted by Newton's formula and the experimental value.
2. Experimental value is 16% greater than the value given by the formula. Newton failed to provide a satisfactory explanation for the difference.
In simple words: Newton's formula for sound velocity in air assumed an isothermal process, but this led to an incorrect value when compared to experimental results, indicating a limitation in its initial assumption.

🎯 Exam Tip: Understanding the assumptions made by Newton and the resulting discrepancies with experimental values is crucial for explaining the limitations of his formula.

3. Solve The Following Problems.

Question 1. A certain sound wave in air has a speed 340 m/s and wavelength 1.7 m for this wave, calculate
(i) the frequency
(ii) the period.
Answer:
Given: v = 340 m/s, λ = 1.7 m
To find: frequency (n), period (T)
Formulae:
i. n = \( \frac{v}{\lambda} \)
ii. T = \( \frac{1}{n} \)
Calculation: From formula, (i)
n = \( \frac{340}{1.7} \)
∴ n = 200 Hz
From formula, (ii)
T = \( \frac{1}{n} \) = \( \frac{1}{2 \times 10^2} \)
= 5 × 10^-3
...... (using reciprocal Table)
∴ T = 0.005 s
In simple words: Given the speed and wavelength of a sound wave, we can calculate its frequency using the wave speed formula and then find its period as the reciprocal of the frequency.

🎯 Exam Tip: Remember the fundamental wave equations \( v = n\lambda \) and \( T = 1/n \) for quick calculations in wave problems.

Question 2. A tuning fork of frequency 170 Hz produces sound waves of wavelength 2m. Calculate speed of sound.
Answer:
Given: n = 170 Hz, λ = 2 m
To find: velocity of sound (v)
Formula: v = nλ
Calculation: From formula,
v = 170 x 2
∴ v = 340 m/s
In simple words: The speed of sound is found by multiplying its frequency by its wavelength, a direct application of the wave equation.

🎯 Exam Tip: Ensure units are consistent (Hz, m, m/s) when applying the wave equation \( v = n\lambda \) to avoid errors.

Question 3. An echo-sounder in a fishing boat receives an echo from a shoal of fish 0.45s after it was sent. If the speed of sound in water is 1500 m/s, how deep is the shoal?
Answer:
Given: t = 0.45 s, v = 1500 m/s,
To Find: depth (d)
Formula: speed (v) = \( \frac{\text{distance}}{\text{time}} \)
Calculation:
For an echo distance travelled by the sound wave = 2 × (distance between echo sounder and shoal) (d)
V = \( \frac{2d}{t} \)
∴ d = \( \frac{1500 \times 0.45}{2} \)
= 337.5 m
In simple words: To find the depth using an echo, we calculate the total distance sound traveled (speed × time) and then divide it by two, as the sound traveled to the shoal and back.

🎯 Exam Tip: For echo problems, remember that the total distance traveled by sound is twice the actual distance to the object, so divide the calculated distance by 2 for the final answer.

Question 4. A girl stands 170 m away from a high wall and claps her hands at a steady rateso that each clap coincides with the echo of the one before.
a) If she makes 60 claps in 1 minute, what value should be the speed of sound in air?
b) Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.
Answer:
i. When the girl makes 60 claps in 1 minute, the value of speed of is 340 m/s.
ii. The girl is at a distance of 226.67 m from the wall when she produces 45 claps per minute.
[Note: The answer given above is calculated in accordance with textual method considering the given data]
In simple words: When claps coincide with echoes, the time between claps equals the time for sound to travel to the wall and back, allowing calculation of sound speed or distance based on the clapping rate.

🎯 Exam Tip: In echo-synchronization problems, the time taken for one clap-echo cycle is key. For part (a), calculate time per clap, then use distance = speed × time (where time is for a round trip). For part (b), use the calculated speed to find the new distance.

Question 5. Sound wave A has period 0.015 s, sound wave B has period 0.025. Which sound has greater frequency?
Answer:
Given: T_A = 0.015 s, T_B = 0.025 s
To find: greater frequency (n)
Formula: n = \( \frac{1}{T} \)
Calculation: From formula,
n_A = \( \frac{1}{T_A} \) = \( \frac{1}{0.015} \)
∴ n_A = 66.67
(using reciprocal table)
n_B = \( \frac{1}{T_B} \) = \( \frac{1}{0.025} \)
∴ n_B = 40 Hz
(using reciprocal table)
∴ n_A > n_B
In simple words: Frequency is the reciprocal of the period; a shorter period means a higher frequency, so the sound wave with the smaller period will have a greater frequency.

🎯 Exam Tip: The relationship \( n = 1/T \) is fundamental. A common mistake is forgetting this inverse relationship when comparing frequencies from given periods.

Question 6. At what temperature will the speed of sound in air be 1.75 times its speed at N.T.P?
Answer:
Given:
V_air = 1.75 V_S.T.P = \( \frac{7}{4} \) V_S.T.P
T_S.T.P = 273 K
To find: temperature T_air
Formula: v \( \propto \sqrt{T} \)
Calculation: From formula,
\( \frac{V_{STP}}{V_{air}} = \sqrt{\frac{T_{STP}}{T_{air}}} \)
∴ \( \frac{V_{STP}}{1.75 V_{air}} = \sqrt{\frac{273}{T_{air}}} \)
∴ \( \frac{V_{STP}}{\frac{7}{4} V_{STP}} = \sqrt{\frac{273}{T_{air}}} \)
∴ \( \frac{4}{7} = \sqrt{\frac{273}{T_{air}}} \)
∴ \( \frac{16}{49} = \frac{273}{T_{air}} \)
∴ T_air = \( \frac{273 \times 49}{16} \)
T_air = 836 K = 563 °C
In simple words: The speed of sound is directly proportional to the square root of the absolute temperature; to find the temperature at which sound speed increases by a factor, we square that factor and multiply by the initial absolute temperature.

🎯 Exam Tip: Remember to use absolute temperature (Kelvin) when dealing with problems involving the variation of sound speed with temperature. Squaring the ratio of velocities and multiplying by the initial temperature in Kelvin is key.

Question 7. A man standing between 2 parallel cliffs fires a gun. He hearns two echos one after 3 seconds and other after 5 seconds. The separation between the two cliffs is 1360 m, what is the speed of sound?
Answer:
distance (s) = 1360 m,
time for first echo = 3 s,
time for second echo = 5 s
To Find : speed of sound (v)
Formula : speed = \( \frac{\text{distance}}{\text{time}} \)
Calculation:
Time for first echo = 3 s
∴ time taken by sound to travel given distance t₁
= \( \frac{3}{2} \) = 1.5 s
Time for second echo = 5 s
∴ time taken by sound to travel given distance t₂
= \( \frac{5}{2} \) = 2.5 s
∴Total time taken by sound to travel given distance, T = 1.5 + 2.5 = 4 s
From formula,
V = \( \frac{1360}{4} \)
∴v = 340 m/s
In simple words: By finding the time it takes for the sound to travel to each cliff and back (half the echo time) and knowing the total distance between cliffs, we can sum the half-times and divide the total distance by this total time to find the speed of sound.

🎯 Exam Tip: For problems with echoes from two surfaces, it's critical to realize that the sum of the half-times (time to cliff and back, divided by two) should relate to the total distance between the cliffs. This allows for the calculation of the speed of sound.

Question 8. If the velocity of sound in air at a given place on two different days of a given week are in the ratio of 1 : 1.1. Assuming the temperatures on the two days to be same what quantitative conclusion can your draw about the condition on the two days?
Answer:
Let v₁ and v₂ be the velocity of sound on day 1 and day 2 respectively.
\( \frac{v_1}{v_2} = \frac{1}{1.1} \)
We know, v \( \propto \frac{1}{\sqrt{\rho}} \)
Let p₁ and p₂ be the density of air on day 1 and day 2 respectively.
∴ \( \sqrt{\frac{\rho_2}{\rho_1}} = \frac{1}{1.1} \)
\( \frac{\rho_2}{\rho_1} = (\frac{1}{1.1})^2 \)
ρ₁ = 1.1² ρ₂ = 1.21 ρ₂
From above equation, we can conclude,
ρ₁ > ρ₂
∴ V₂ > V₁ i.e., the velocity of sound is greater on the second day than on the first day.
We know, speed of sound in moist air (v_m) is greater than speed of sound in dry air (v_d).
∴ We can conclude, air is moist on second day and dry on the first day.
In simple words: Since sound travels faster in moist air and velocity is inversely proportional to the square root of density, a higher velocity on the second day implies lower air density, suggesting the air was more humid.

🎯 Exam Tip: Remember the inverse relationship between sound velocity and the square root of air density, and that moist air is less dense than dry air, leading to a higher speed of sound.

Question 9. A police car travels towards a stationary observer at a speed of 15 m/s. The siren on the car emits a sound of frequency 250 Hz. Calculate the recorded frequency. The speed of sound is 340 m/s.
Answer:
Given: v_s = 15 m/s, n₀ = 250 Hz, v = 340 m/s
To find: Frequency (n)
Formula: n = n₀(\( \frac{v}{v - v_s} \))
Calculation: As the source approaches listener, apparent frequency is given by,
n = 250 (\( \frac{340}{340 - 15} \)) = 250 (\( \frac{340}{325} \))
n = \( \frac{3400}{13} \)
∴ n = 261.54 Hz
In simple words: When a sound source approaches a stationary observer, the apparent frequency heard by the observer increases due to the Doppler effect, calculated using the source's speed relative to the speed of sound.

🎯 Exam Tip: For Doppler effect problems, correctly identify whether the source or observer is moving, and choose the appropriate signs in the formula for approaching or receding motion. Pay close attention to the formula: n = n₀(\( \frac{v \pm v_L}{v \mp v_S} \)).

Question 10. The sound emitted from the siren of an ambulance has frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance initially travels towards and then away from the observer at a speed of 30 m/s.
Answer:
Given: v_s = 30 m/s, n₀ = 1500 Hz, v = 340 m/s
To find: Difference in apparent frequencies (n_A – n'_A)
Formulae:
i. When the ambulance moves towards he stationary observer then n_A = n₀(\( \frac{v}{v - v_s} \))
ii. When the ambulance moves away from the stationary observer then, n'_A = n₀(\( \frac{v}{v + v_s} \))
Calculation:
From formula (i), icon' 340
n_A = 1500(\( \frac{340}{340 - 30} \))
∴ n_A = 1645 Hz
From (ii)
n'_A = 1500(\( \frac{340}{340 + 30} \))
∴ n'_A = 1378 Hz
Difference between n_A and n'_A
= n_A – n'_A = 1645 – 1378 = 267 Hz
In simple words: The Doppler effect causes the apparent frequency to increase when a sound source approaches and decrease when it recedes; the difference between these two frequencies highlights the total shift experienced by a stationary observer.

🎯 Exam Tip: In Doppler effect problems involving both approaching and receding scenarios, carefully apply the correct signs in the frequency formula. The difference in apparent frequencies is a common calculation that tests understanding of both cases.

Can You Recall? (Textbook Page No. 142)

Question i. What type of wave is a sound wave?
Answer:
i. Sound wave is a longitudinal wave.
In simple words: A sound wave is a longitudinal wave because the particles of the medium vibrate parallel to the direction in which the wave travels.

🎯 Exam Tip: Classifying waves based on particle vibration relative to propagation direction is a fundamental concept. Longitudinal waves involve parallel vibration, like sound.

Question ii. Can sound travel in vacuum?
Answer:
ii. Sound cannot travel in vacuum.
In simple words: Sound requires a medium to propagate because it travels as vibrations of particles, and a vacuum lacks these particles.

🎯 Exam Tip: This is a common conceptual question. Emphasize that sound is a mechanical wave requiring a medium, unlike electromagnetic waves.

Question iii. What are reverberation and echo?
Answer:
iii. a. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
b. An echo is the repetition of the original sound because of reflection by some surface.
In simple words: Reverberation is the persistence of sound due to multiple reflections within an enclosed space, while an echo is a distinct repetition of a sound due to a single, delayed reflection.

🎯 Exam Tip: Distinguishing between reverberation (multiple reflections, sound persists) and echo (distinct, delayed reflection) is important, often tested with examples or definitions.

Question iv. What is meant by pitch of a sound?
Answer:
iv. The characteristic of sound which is determined by the value of frequency is called as the pitch of the sound.
In simple words: Pitch is the perceptual quality of a sound that allows listeners to order them on a musical scale, primarily determined by the frequency of the sound wave.

🎯 Exam Tip: Associate pitch directly with frequency. Higher frequency means higher pitch, and vice-versa. This is a basic but essential concept in sound properties.

Activity (Textbook Page No. 144)

Question i. Using axes of displacement and distance, sketch two waves A and B such that A has twice the wavelength and half the amplitude of B.
Answer:
In simple words: To sketch waves with specific wavelength and amplitude ratios, visually represent wave A as having broader cycles and half the peak height compared to wave B, ensuring the peaks and troughs align with the given conditions.

🎯 Exam Tip: When sketching waves, accurately represent amplitude (maximum displacement from equilibrium) and wavelength (distance for one complete cycle) based on the given ratios. Labeling axes is also crucial.

Question ii. Determine the wavelength and amplitude of each of the two waves P and Q shown in figure below.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विस्थापन (Y-अक्ष) और दूरी (X-अक्ष) के बीच दो तरंगों A और B को दर्शाता है। तरंग A का आयाम 2 मीटर है और इसकी तरंगदैर्ध्य 4 मीटर है। तरंग B का आयाम 4 मीटर है और इसकी तरंगदैर्ध्य 2 मीटर है। X-अक्ष पर 1 सेमी = 1 मीटर और Y-अक्ष पर 1 सेमी = 1 मीटर का पैमाना दिया गया है।

i. In phase point: A and F; B and H; C and I; D and J
ii. Out of phase points: A and B, B and D, Fl and J, E and F,
iii. Point having phase difference of π/2: A and B; B and C; D; D and F; F and H; H and I; J and I

In simple words: To determine wavelength and amplitude from a wave diagram, measure the horizontal distance for one complete cycle (wavelength) and the maximum vertical displacement from the equilibrium position (amplitude).

🎯 Exam Tip: Practice reading wave diagrams to accurately identify wavelength (peak-to-peak or trough-to-trough distance) and amplitude (half the peak-to-trough vertical distance from equilibrium) using the given scales.