Q: Question 16The speed of sound in a gas is given by :V = Pγρ\sqrt\dfrac{\text P}{\text {γρ}}γρPV = γPρ\sqrt\dfrac{\text {γP}}{\text ρ}ργPV = γρP\sqrt{\dfrac{\gamma \text ρ}{\text P}}PγρV = γρP\sqrt{\text {γρP}}γρP

V = γPρ\sqrt\dfrac{\text {γP}}{\text ρ}ργPReason — According to Laplace, when sound travels in a gas, during the formation of compression and rarefaction, there is no exchange of heat in the medium i.e., the propagation of sound is an adiabatic change.Hence, the speed of sound in a gas is given by γPρ\sqrt\dfrac{\text {γP}}{\textρ}ργP

Question 1

Question 6Describe a simple experiment to determine the speed of sound in air. What approximation is made in the method described by you?

Accepted Answer

The fact that light travels in air about a million times faster than sound, can be used to determine the speed of sound in air.Experiment —Choose two places A and B at high altitudes facing facing each other, at a distance d apart (say, about 1 km), in still iair. The distance d is noted. At each place, there is an observer with a gun and a stop watch. First the observer at place A fires the gun, while the observer at place B starts his stop watch immediately on seeing the flash of fire at A and stops it when he hears the sound of fire. The observer at B by his watch thus finds the time interval t1 taken by the sound to travel from A to B.Now the observer at place B fires the gun. The observer at A starts his stop watch when he sees the flash of fire at B and stops when he hears the sound of fire. Thus, the observer A by his watch finds the time interval t2 taken by sound to travel from B to A.The average of the two time intervals is t = t1+t22\dfrac{	ext t_1 + 	ext t_2 }{2}2t1​+t2​​. This is the time taken by sound to travel the distance d between the places A and B.The speed of sound is calculated by using the formula —V = DistanceTime\dfrac{	ext{Distance}}{	ext{Time}}TimeDistance​ = dt\dfrac{	ext d}{	ext t}td​ m s-1In the experiment the speed of sound determined is not very accurate because of the personal error of the two observers and the variation in temperature and humidity of air in between the places A and B.

Question 2

Question 16The speed of sound in a gas is given by :V = Pγρ\sqrt\dfrac{	ext P}{	ext {γρ}}γρP​​V = γPρ\sqrt\dfrac{	ext {γP}}{	ext ρ}ργP​​V = γρP\sqrt{\dfrac{\gamma 	ext ρ}{	ext P}}Pγρ​​V = γρP\sqrt{	ext {γρP}}γρP​

Accepted Answer

V = γPρ\sqrt\dfrac{	ext {γP}}{	ext ρ}ργP​​Reason — According to Laplace, when sound travels in a gas, during the formation of compression and rarefaction, there is no exchange of heat in the medium i.e., the propagation of sound is an adiabatic change.Hence, the speed of sound in a gas is given by γPρ\sqrt\dfrac{	ext {γP}}{	extρ}ργP​​

Question 3

Question 17The speed of sound in air at 0°C is nearly :1450 m s-1450 m s-15100 m s-1330 m s-1

Accepted Answer

330 m s-1Reason — The speed of sound in air at 0°C is 330 m s-1

Question 4

Question 18The speed of sound in a gas is related to temperature (in Kelvin) as :directly proportional to temperature.directly proportional to the square root of temperature.inversely proportional to temperature.inversely proportional to the square root of temperature.

Accepted Answer

directly proportional to the square root of temperature.Reason — The speed of sound in a gas increases with the increase in temperature of the gas. The reason is that with the increase in temperature, the density of the gas decreases and consequently the speed of sound increases. The speed of sound in a gas is directly proportional to the square root of temperature.

Question 5

Question 19Factors affecting the speed of sound are :temperaturedensitypressureonly (a) and (b)

Accepted Answer

only (a) and (b)Reason — Factors affecting the speed of sound are:densitytemperaturehumidity anddirection of wind.

Question 6

Question 1The heart of a man beats 75 times a minute. What is it's (a) frequency and (b) time period?

Accepted Answer

(a) Given,Heart beats 75 times a minuteConverting minute into seconds, we get,1 min = 60 sAs, frequency (f) is the number of times the heart beats in a secondTherefore, we get,f=7560f=1.25 s−1\text f = \dfrac{75}{60} \[0.5em] \text f = 1.25 \text{ s}^{-1}f=6075​f=1.25 s−1Hence, Frequency = 1.25 s-1(b) Time period t = 1f\dfrac{1}{\text f}f1​Substituting the values, we get,t=11.25t=0.8 s\text t = \dfrac{1}{1.25} \[0.5em] \text t = 0.8\ st=1.251​t=0.8 sHence, t = 0.8 s

Question 7

Question 2The time period of a simple pendulum is 2 s. Find it's frequency.

Accepted Answer

Given,time (t) = 2 sfrequency = ?As,Frequency (f) = 1t\dfrac{1}{\text t}t1​Substituting the values, we get,f=12f=0.5 Hz\text f = \dfrac{1}{2} \[0.5em] \text f = 0.5 \text{ Hz}f=21​f=0.5 HzHence, f = 0.5 Hz

Question 8

Question 3The separation between two consecutive crests in a transverse wave is 100 m. If wave velocity is 20 m s-1, find the frequency of wave.

Accepted Answer

Given,wavelength (λ) = 100 mwave velocity (V)= 20 m s-1Frequency (f) = ?As we know,Wave velocity (V) = Frequency (f) x Wavelength (λ)Substituting the values, we get,20=f×100⇒f=20100f=0.2 Hz20 = \text f \times 100 \[0.5em] \Rightarrow \text f = \dfrac{20}{100} \[0.5em] \text f = 0.2 \text{ Hz}20=f×100⇒f=10020​f=0.2 HzHence, f = 0.2 Hz

Question 9

Question 4A longitudinal wave travels at a speed of 0.3 m s-1 and the frequency of a wave is 20 Hz. Find the separation between the two consecutive compressions.

Accepted Answer

Given,Wave velocity (V)= 0.3 m s-1frequency (f) = 20 Hzwavelength (λ) = ?As we know,Wave velocity (V) = Frequency (f) x Wavelength (λ)Substituting the values, we get,0.3=20×λ⇒λ=0.320⇒λ=3200⇒λ=1.5×10−2 m0.3 = 20 \times \text λ \[0.5em] \Rightarrow \text λ = \dfrac{0.3}{20} \[0.5em] \Rightarrow \text λ = \dfrac{3}{200} \[0.5em] \Rightarrow \text λ = 1.5 \times 10^{-2} \text { m}\[0.5em]0.3=20×λ⇒λ=200.3​⇒λ=2003​⇒λ=1.5×10−2 mHence, the separation between the two consecutive compressions which is the wave length = 1.5 x 10-2 m ( or 1.5 cm)

Question 10

Question 5A source of wave produces 40 crests and 40 troughs in 0.4 s. What is the frequency of the wave?

Accepted Answer

Given,One crest and one trough make one wave, hence there are 40 waves.time (t) = 0.4 sfrequency f = ?Frequency is the number of waves per second,f=400.4⇒f=100 Hz\text f = \dfrac{40}{0.4} \[0.5em] \Rightarrow \text f = 100 \text { Hz}f=0.440​⇒f=100 HzHence,Frequency of the wave = 100 Hz

Question 11

Question 6An observer A fires a gun and another observer B at a distance 1650 m away from A hears it's sound. If the speed of sound is 330 m s-1, find the time when B will hear the sound after firing by A.

Accepted Answer

Given,Distance between A and B = 1650 mSpeed (V)= 330 m s-1time t = ?V = Distance (d)Time (t)\dfrac{\text{Distance (d)}}{\text {Time (t)}}Time (t)Distance (d)​Substituting the values, we get,330=1650t⇒t=1650330⇒t=5 s330 = \dfrac{1650}{\text t} \[0.5em] \Rightarrow \text t = \dfrac{1650}{330} \[0.5em] \Rightarrow \text t = 5 \text { s} \[0.5em]330=t1650​⇒t=3301650​⇒t=5 sTherefore, the time when B will hear the sound after firing by A = 5 s.

Question 12

Question 8A boy fires a gun and another boy at a distance hears the sound of fire 2.5 s after seeing the flash. If speed of sound in air is 340 m s-1 , find the distance between the boys.

Accepted Answer

Given,Time (t)= 2.5 sSpeed of sound (V) = 340 m s-1Distance (d) = ?V = Distance (d)Time (t)\dfrac{\text{Distance (d)}}{\text {Time (t)}}Time (t)Distance (d)​Substituting the values, we get,340=d2.5⇒d=340×2.5⇒d=850 m340 = \dfrac{\text d}{2.5} \[0.5em] \Rightarrow \text d = 340 \times 2.5 \[0.5em] \Rightarrow \text d = 850 \text { m} \[0.5em]340=2.5d​⇒d=340×2.5⇒d=850 mTherefore, the distance between the boys = 850 m.

Question 13

Question 9An observer sitting in line of two tanks, watches the flashes of two tanks firing at each other at the same time, but he hears the sounds of two shots 2s and 3.5s after seeing the flashes. If distance between the two tanks is 510 m, find the speed of sound.

Accepted Answer

Given,Time taken to hear the sound of tank 1 = 2 sTime taken to hear the sound of tank 2 = 3.5 sTime interval = 3.5 – 2 = 1.5 sDistance between two tanks = 510 mVelocity (V) = ?V = Distance (d)Time (t)\dfrac{\text{Distance (d)}}{\text {Time (t)}}Time (t)Distance (d)​Substituting the values, we get,V=5101.5⇒V=340 m s−1\text V = \dfrac{510}{1.5} \[0.5em] \Rightarrow \text V = 340 \text { m s}^{-1}\[0.5em]V=1.5510​⇒V=340 m s−1Therefore, the speed of sound = 340 m s-1 .

Question 14

Question 10How long will sound take to travel in (a) an iron rail and (b) air, both 3.3 km in length? Take speed of sound in air to be 330 m s-1 and in iron to be 5280 m s-1 .

Accepted Answer

(a) Given,Length of the iron rail = 3.3 kmConverting km to m, we get,1 km = 1000 mTherefore, 3.3 km = 1000 x 3.3 = 3300 mSpeed of the sound in iron (V) = 5280 m s-1t = ?V = Distance (d)Time (t)\dfrac{\text{Distance (d)}}{\text {Time (t)}}Time (t)Distance (d)​Substituting the values, we get,5280=3300t⇒t=33005280⇒t=0.625 s5280 = \dfrac{3300}{\text t} \[0.5em] \Rightarrow \text t = \dfrac{3300}{5280} \[0.5em] \Rightarrow \text t = 0.625 \text { s} \[0.5em]5280=t3300​⇒t=52803300​⇒t=0.625 sTherefore, the time taken by sound to travel in iron rail = 0.625 s(b) Speed of sound in air = 330 m s-1Substituting the values in the above formula, we get,330=3300t⇒t=3300330⇒t=10 s330 = \dfrac{3300}{\text t} \[0.5em] \Rightarrow \text t = \dfrac{3300}{330} \[0.5em] \Rightarrow \text t = 10 \text { s} \[0.5em]330=t3300​⇒t=3303300​⇒t=10 sTherefore, the time taken by sound to travel in air = 10 s

Question 15

Question 11Assuming the speed of sound in air equal to 340 m s-1 and in water equal to 1360 m s-1 , find the time taken to travel a distance 1700 m by sound in (i) air and (ii) water.

Accepted Answer

(i) Given,Speed of sound in air (Va) = 340 m s-1Speed of sound in water (Vw) = 1360 m s-1Distance (d) = 1700 mt = ?V = Distance (d)Time (t)\dfrac{\text{Distance (d)}}{\text {Time (t)}}Time (t)Distance (d)​Substituting the values for air, we get,340=1700t⇒t=1700340⇒t=5 s340 = \dfrac{1700}{\text t} \[0.5em] \Rightarrow \text t = \dfrac{1700}{340} \[0.5em] \Rightarrow \text t = 5 \text { s} \[0.5em]340=t1700​⇒t=3401700​⇒t=5 sTherefore, the time taken by sound to travel in air = 5 s(ii) Substituting the values for water, we get,1360=1700t⇒t=17001360⇒t=1.25 s1360 = \dfrac{1700}{\text t} \[0.5em] \Rightarrow \text t = \dfrac{1700}{1360} \[0.5em] \Rightarrow \text t = 1.25 \text { s} \[0.5em]1360=t1700​⇒t=13601700​⇒t=1.25 sTherefore, the time taken by sound to travel in air = 1.25 s

Question 16

Question 13In which medium the speed of sound is more — humid air or dry air? Give a reason to your answer.

Accepted Answer

The speed of sound is more in humid air.From the relation.V ∝ γPρ\sqrt\dfrac{	ext {γP}}{	ext ρ}ργP​​, it is clear that V ∝ 1ρ\dfrac{1}{\sqrt	ext ρ}ρ​1​ i.e., the speed of sound is inversely proportional to the square root of density of the gas.As the density of air decreases with an increase in moisture level in air hence speed of sound increases in humid air.

Question 17

Question 12Answer the following —(a) Can sound travel in vacuum?(b) How does the speed of sound differ in different media?

Accepted Answer

(a) No, sound cannot travel in vacuum because sound needs a medium to propagate and there are no particles in vacuum to propagate sound.(b) The speed of sound differs in different media. The speed of sound is more in solids, less in liquids and least in gases (since solids are much more elastic than liquids and gases). The speed of sound is nearly 5100 m s-1 in steel, 1450 m s-1 in water and 330 m s-1 in air at 0°C.

Question 18

Question 14How does the speed of sound in air vary with temperature?

Accepted Answer

The speed of sound is directly proportional to the square root of temperature of the medium. It increases by about 0.61 m s-1 (or 61 cm per second) for each 1°C rise in temperature (provided that the rise in temperature is not very large).i.e.,Vt = V0 + 0.61t

Question 19

Question 15Complete the following sentences —(a) Sound cannot travel through ............... ; it requires a ...............(b) When sound travels in a medium, the particles of medium ............... but the disturbance...............(c) A longitudinal wave is composed of compression and ...............(d) A transverse wave is composed of crest and ...............(e) Wave velocity = ............... x wavelength

Accepted Answer

(a) Sound cannot travel through vacuum ; it requires a medium.(b) When sound travels in a medium, the particles of medium do not move but the disturbance moves ahead.(c) A longitudinal wave is composed of compression and rarefaction.(d) A transverse wave is composed of crest and trough.(e) Wave velocity = frequency x wavelength

Propagation of Sound Waves

Exercise 8A Long Answer Type

Exercise 8A Multiple Choice Type

Exercise 8A Numericals

Exercise 8A Short Answer Type

Exercise 8A Very Short Answer Type