Solutions-Class-11-Science-Physics-Chapter-8-Sound-Maharashtra Board

(D) Longitudinal progressive

(B) Frequency

(C) propagate adiabatically

(D) solid

(D) absorb sound

• When a wave travels through a medium, at any point of the medium, the form of the wave repeats after equal intervals of time. Thus, a wave motion is periodic in time.
• At any instant, the form of the wave repeats at equal distances along the direction of wave propagation. Thus, a wave motion is periodic in space.

Thus, a wave motion is doubly periodic.

The apparent change in the frequency of sound heard by a listener, due to the relative motion of the source of sound and the listener, is called the Doppler effect in sound.

A progressive wave in which the vibration of the individual particles of the medium is

perpendicular to the direction of propagation of the wave is called a transverse progressive wave.

Characteristics of a transverse wave:

• Particles vibrate perpendicular to wave propagation direction with same period and amplitude.
• Progressively changing phase of particles forms alternate crests and troughs, forming one cycle.
• No pressure and density variations at any medium point during wave passage.
• Propagation depends on medium's rigidity, requiring modulus of rigidity.
• Waves can only travel through solids and liquid surfaces, not through liquids and gases.
• Vibrations of medium particles lie in the same plane, exhibiting polarization.

A progressive wave in which the vibration of the individual particles of the medium is along the direction of propagation of the wave is called a longitudinal progressive wave.

Examples:

• A long spring (slinky) is attached at one end to a simple oscillatory mechanism, as shown in Fig. (a). When the metal ball is displaced slightly to one side and released, its oscillations periodically compress and extend the spring. As a result, a wave motion propagates along the length of the spring. Since each coil of the spring oscillates along the length of the spring, the wave motion is longitudinal.
• If a long metal bar is periodically struck on one end, Fig. (b), each impact compresses the material of the bar followed by an expansion. The particles of the bar move back and forth and thus a longitudinal wave travels along the bar.

Newton's formula for velocity of sound:

v = \(\sqrt{\frac{K}{p}}\)

where v is the wave speed, and p and K are respectively the density and the bulk modulus of the medium.

According to Laplace's formula, the speed of sound in a gas is

v = \(\sqrt{\frac{γp}{ρ}}\)    …..(1)

where p and ρ are the pressure and density of the gas and

ν = C_P/C_V

In terms of the density of the gas, the equation of state for an ideal gas is written as

\(\frac{p}{ρT}=\frac{R}{M}\) = constant    ….(2)

where T is the absolute temperature, R is the universal gas constant and M is the molar mass.

Therefore, at constant T,

\(\frac{p}{ρ}\) = constant     …..(3)

Hence, v = \(\sqrt{\frac{γp}{ρ}}\) = constant  ……(4)

Thus, at constant temperature, the speed of sound in a gas is independent of the pressure provided the pressure is atmospheric or below, so that the gas closely approximates an ideal gas.

The term humidity is used to express the moisture content in the air. The presence of water vapour in the air affects the density of air which in turn affects the speed of sound in air.

Let v_m be the speed of sound in moist air having density ρ_m and v_d the speed of sound in dry air having density ρ_d.

According to Laplace's formula, v_m = \(\sqrt{\frac{γp}{ρ_m}}\) and v_d = \(\sqrt{\frac{γp}{ρ_d}}\)

where p is the pressure of air and ν = C_P/C_V

∴ At constant pressure, \(\frac{v_m}{v_d}=\sqrt{\frac{ρ_d}{ρ_m}}\)

The density of moist air is less than that of dry air.

v_m > v_d

That is, the speed of sound in moist air is greater than that in dry air.

• An echo is a repetition of the original sound due to reflection from a rigid surface at a distance from the source.
• It can be heard after a time interval in hilly regions or near a reflecting medium.
• Echoes can be heard multiple times due to multiple reflections, but a true echo is a single reflection.

• Bats use echolocation to navigate and find food in the dark. They emit short ultrasonic pulses (frequency range : 30 kHz to 150 khz). Their echoes give the bats information about obstacles and food.
• Dolphins use echolocation for underwater navigation, location of prey and protection from predators, using subsonic pulses of frequency about 100 Hz.

• Amplitude : The magnitude of the maximum displacement of a particle of the medium from its equilibrium position is called the amplitude of the wave.
• Wavelength : It is the distance between consecutive particles of the medium which are moving in exactly the same way at the same time and have the same displacement from their equilibrium positions.

(i) Points A and E, C and G, B and F, D and H are in phase.

(ii) Points A and C, B and D, C and E, D and F,  E and G, F and H, are out of phase.

(iii) Points A and B, B and C, C and D, D and E, E and F, F and G, G and H have a phase difference of /2 radian.

During the passage of a progressive wave through a medium, the wave advances a

distance equal to the wavelength λ. in the periodic time T of the vibration of the particles of the medium. Then, the speed of propagation (the magnitude of the velocity) of the wave is

v = λ/T

The frequency n of the progressive wave is equal to the reciprocal of the period :

n = 1/T  (2)

From Eqs. (1) and (2), we have,

v = n λ . i.e., wave speed = frequency x wavelength

This relation is true for all types of progressive wave.

Principle of superposition of waves: When two or more waves arrive at the same time at a point, the resultant displacement at that point, at that instant, is the vector sum of the displacements due to each wave acting independently.

Explanation: Consider two waves travelling in a medium arriving simultaneously at a point P.

Let y₁ and y₂ be the displacements due to the two waves at the point. Then, according to the principle of superposition of waves, the resultant displacement at the point is \(\vec{y_R}=\vec{y_1}+\vec{y_2}\)

This implies that, depending upon the phase difference between the two waves,

y_R (maximum) = |y₁| + |y₂ | and

y_R (minimum) = |y₁| — |y₂ |

Let n ≡ apparent frequency of sound,

n₀ ≡ actual frequency of sound,

v ≡ speed of sound in air,

v ≡ speed of the source of sound and

vL ≡ speed of the listener

(i) For the source and listener moving towards (approaching) each other,

n = n₀ \([\frac{v+v_L}{v-v_s}]\) 

(ii) For the source and listener moving away (receding) from each other,

n = n₀ \([\frac{v-v_L}{v+v_s}]\)

Let n ≡ apparent frequency of sound,

n₀ ≡ actual frequency of sound

v ≡ speed of sound in air and

v_L ≡ speed of the source of sound

(i) The apparent frequency when the source is moving towards (approaching) the stationary listener is

n = n₀ \([\frac{v+v_L}{v}]\)

(ii) The apparent frequency when the source is moving away (receding) from the stationary listener is

n = n₀ \([\frac{v-v_L}{v}]\)

Let n ≡ apparent frequency of sound,

n₀ ≡ actual frequency of sound

v ≡ speed of sound in air and

v_s ≡ speed of the source of sound

(i) The apparent frequency when the source is moving towards (approaching) the stationary listener is

n = n₀ \([\frac{v}{v-v_s}]\)

(ii) The apparent frequency when the source is moving away (receding) from the stationary listener is

n = n₀ \([\frac{v}{v+v_s}]\)

Let us consider propagation of a mechanical wave. We shall consider a simple case where the wave travels along the positive x-axis and the displacement (y) of a particle of the medium from its equilibrium position is given by

y(x,t) = a sin\((\frac{2π}{λ}x+2πnt+Φ)\)

Here, a is the amplitude, A is the wavelength and n is the frequency of wave motion. x denotes the equilibrium position of the particle and t denotes time.  is a constant. y may be along the x axis or perpendicular to it.

In this case, the state of oscillation of a particle of the medium is called the phase of the particle and also of the wave.

The quantity \((\frac{2π}{λ}x+2πnt+Φ)\) is the phase of the particle at time t.

To describe the phase of a particle at a place, its displacement, direction of velocity and oscillation number are required.

In above Fig. particles O and E or Q and L or H and D or P and K or G and C, etc. have the same displacement I as well as direction of velocity. These particles are said to be in phase during their respective oscillations. Particles O and G, Q and P, have the same displacement but the directions of their velocity are opposite. These particles are said to be out of phase during their respective oscillations.

Successive particles having the same phase differ by 1 in their oscillation number. Thus, if particle F is at its n^th oscillation, particle B is at its (n + 1)^th oscillation.

(a) Progressive wave : A wave motion or a progressive wave is a state of disturbance in which energy is propagated from one place to another at a finite speed without propagation of matter. It is also called a travelling wave.

(b) Properties of a progressive wave : Amplitude, wavelength, period, double periodicity, frequency and wave speed are the properties of a progressive wave.

(a) Newton (in 1687) derived the relation between the speed of sound, and the inertia and the elastic property of the medium:

v = \(\sqrt{\frac{K}{ρ}}\)

where v is the wave speed ‘(speed of sound) and ρ and K are respectively the density and the bulk modulus of the medium.

Assumptions: For a theoretical estimate of the speed of sound in air, Newton assumed that the air obeyed Boyle's law and that the pressure variations caused by the passage of the waves are slow and take place isothermally. Newton's formula: Newton, therefore, used the isothermal bulk modulus, which, for an ideal gas, is equal to the pressure p of the gas. So, the expression he deduced for the speed of sound in air is

v = \(\sqrt{\frac{K_{isothermal}}{ρ}}\) = \(\sqrt{\frac{p}{ρ}}\)

This is Newton's formula for the speed of sound in an ideal gas.

(b) At NTP, p = 1.013 x 10⁵ Pa and

p_air = 1.293 kg/m³, so that

v = \(\sqrt{\frac{1.013×10^5\,Pa}{1.293\,kg/m^3}}\) = 280 m/s

Limitation: At NTP, the value of speed of sound given by Newton's formula (280 m/s) is significantly less (by about 16%) than the experimental value of 331.3 m/s.

Given: v = 340 m/s, λ = 1.7 m

(1) The frequency of the wave, n = \(\frac{v}{λ}\) = \(\frac{340}{λ}\) = 200 Hz

(2) The period of the wave, T = \(\frac{1}{n}\) = \(\frac{1}{200}\) = 5 x 10^-3 s

Given : n =170 Hz, λ = 2 m 

The speed of sound, v = nλ = 170 x 2 = 340 m/s

Given : Total time, 2t = 0.45 s  ∴ t = 0.225 s, v = 1500 m/s

s = v t = (1500) (0.225) = 337.5 m

The shoal is 337.5 m deep.

(a) 60 claps in 1 minute, ∴ 1 clap in 1 second

In Fig.

S = GW = 170 m

v = \(\frac{distance}{time} = \frac{GW+WG}{time}\) = \(\frac{170+170}{1}\) = 340 m/S

This is the speed of sound in air.

(b) Let G’ correspond to the location in part (b)

Her 45 claps in 1 minute ∴ 1 clap corresponds to \(\frac{60}{45}\)s = \(\frac{4}{3}\)s

In this case, G‘W + WG' = 2G‘W = vt

= 340 × \(\frac{4}{3}\) = \(\frac{340×4}{3}\) m

G‘W = \(\frac{340×4}{3×2}\) = \(\frac{340×2}{3}\) = \(\frac{680}{3}\) = 226.7 m

This is the required distance.

(If we take 40 claps in 1 minute then answer = 255m as per textbook)

Given : T_A = 0.015 s, T_B = 0.025 s

Frequency, n = \(\frac{1}{T}\)

Here, T_A < T_B   ∴ n_A > n_B

Sound wave A has greater frequency

Given : v = 1.75v₀, T₀= 273 K

v = \(\sqrt{\frac{rRT}{M}}\) and v₀ = \(\sqrt{\frac{rRT_0}{M}}\)

\(\frac{v}{v_0}\)= \(\sqrt{\frac{T}{T_0}}\)

   ∴ \(\frac{T}{T_0}\) = \((\frac{v}{v_0})^2\)

∴ T = \(T_0(\frac{v}{v_0})^2\)

= 273(1.75)² = 836.1 K

= 836.1 − 273 = 563.1 °c

Given : 2t₁ = 3 s, ∴ t₁ = 1.5 s, 2t₂ = 5 s, ∴  t₂ = 2.5 s,

s₁+ s₂ = 1360 m

s₁ = vt₁ = 1.5 v and s₂ = vt₂ = 2.5v

∴ s₁ + s₂ = 1.5 v+ 2.5 v = 4v

v = \(\frac{s_1+s_2}{4}=\frac{1360}{4}) = 340m/s

340 m/s is the speed of sound in air.

Given :  \(\frac{v_1}{v_2}\) = \(\frac{1}{1.1}\)

v = \(\sqrt{\frac{γp}{ρ}}\),  v₁ = \(\sqrt{\frac{γp}{ρ_1}}\) and v₂ = \(\sqrt{\frac{γp}{ρ_2}}\)

for the same pressure

∴ \(\frac{v_2}{v_1}\) = \(\sqrt{\frac{ρ_1}{ρ_2}}\)

∴ \(\sqrt{\frac{ρ_1}{ρ_2}}\) = (1.1)² = 1.21

Thus, ρ₁ = 1.21ρ₂, where ρ₁ and ρ₂ are the densities of air on the two days.

Given : v_s = 15 m/s, v = 340 m/s, n₀ = 250 Hz

n = n₀\((\frac{v}{v-v_s})\) = (250)\((\frac{340}{340-15})\)

   = \(\frac{250×340}{325}\)  = \(\frac{3400}{13}\) 

   = 261.54 Hz

This is the recorded (apparent) frequency.

Given : n₀ = 1500 Hz, v = 340 m/s, v_s = 30 m/s

n = n₀\((\frac{v}{v-v_s})\)  = (1500)\((\frac{340}{340-30})\)

= \(\frac{1500×340}{310}\) = 1645 Hz

(as the ambulance moves towards the stationary observer)

n = n₀\((\frac{v}{v+v_s})\)  = (1500)\((\frac{340}{340+30})\)

= \(\frac{1500×340}{370}\) = 1378 Hz

(as the ambulance moves away from the stationary observer)

The difference in frequencies

= n₁ − n₂ = 1645 − 1378

= 267 Hz