Standard 9 Science and Technology - 12. Study of Sound

Question 1:

Fill in the blanks and explain.

a. Sound does not travel through ............

b. The velocity of sound in steel is ............... than the velocity of sound in water.

c. The incidence of ......... in daily life shows that the velocity of sound is less than the velocity of light.

d. To discover a sunken ship or objects deep inside the sea, ....................... technology is used.

Answer 1:

a. Sound does not travel through vacuum.

b. The velocity of sound in steel is greater than the velocity of sound in water.

c. The incidence of thunderstorm in daily life shows that the velocity of sound is less than the velocity of light.

d. To discover a sunken ship or objects deep inside the sea, SONAR technology is used.

Question 2:

Explain giving scientific reasons.

a. The roof of a movie theatre and a conference hall is curved.

b. The intensity of reverberation is higher in a closed and empty house.

c. We cannot hear the echo produced in a classroom.

Answer 2:

a. The roof of a movie theatre and a conference hall is curved so that the sound after reflection from the ceiling reaches all the parts of the hall uniformly.

b. In an empty house, there is no sound absorbers present such as furnitures, curtains etc. Also, if the room is closed, the sound cannot escape the room. Thus, the sound goes through multiple reflections from the rigid walls of a closed and empty room without the reflections of the sound being absorbed. Hence, the intensity of reflections do not fade which would have faded if the room was not empty (has sound absorbers). Due to this the intensity of reverberation is higher in a closed and empty room.

c. The classrooms are designed in such a way that the distance between the walls are less than 17.2 m. Due to this, the reflections of sound from to walls or echos reach our ears before 0.1 s. Because of this, we are not able to distinguish between the original sound and the echo produced in the classroom.

Question 3:

Answer the following questions in your own words.

a. What is an echo? What factors are important to get a distinct echo?

b. Study the construction of the Golghumat at Vijapur and discuss the reasons for the multiple echoes produced there.

c. What should be the dimensions and the shape of classrooms so that no echo can be produced there?

Answer 3:

a. The repetition of sound caused by its reflection off a hard surface is known as echo. Factors important to get a distinct echo are:

The minimum distance between the source and reflector of sound should be 17 m.
The size of reflector must be large enough as compared to the wavelength of the sound wave.

b. Golghumbat has a central dome which stands without any support. The sound produced here gets reflected throughout the dome producing the echo at least three to four times in a second. If the surroundings inside the architecture are very quiet, then we would be able to hear echo at least seven to ten times. So, this central dome is the main reason for the production of multiple echoes in Golghumat.

c. For no echo to be produced in a classroom, the classroom should be square shaped with distance between the walls to be less than 17.2 m. Also, the ceiling should be curved shaped so that the reflection of sound wave is uniform throughout the room.

Question 4:

Where and why are sound absorbing materials used?

Answer 4:

The sound absorbing materials are used on the roofs and walls of auditoriums, concert halls, theatres, etc. as well as upholsteries choosen for these rooms are made of sound absorbing materials. In these places, excessive reverberation is highly undesirable. Due to the presence of sound absorbing materials, the sound reflected from the rigid surfaces is absorbed by these materials and hence reverberation is subdued.

Question 5:

Solve the following examples.

a. The speed of sound in air at 0oC is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?

b. Nita heard the sound of lightning after 4 seconds of seeing it. What was the distance of the lightning from her?(The velocity of sound in air is 340 m/s?)

c. Sunil is standing between two walls. The wall closest to him is at a distance of 360 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.

1. What is the velocity of sound in air?

2. What is the distance between the two walls?

d. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gm respectively. In which bottle will sound travel faster? How may times as fast as the other?

e. Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw?

Answer 5:

a. Let the temperature at which the velocity of air becomes 344 m/s be ToC.
The increase in velocity of sound in air from 0oC to ToC = 344

-

332 = 12 m/s
Given,
The velocity of sound in air increases by 0.6 m/s for 1oC rise in temperature.
Therefore, for 1 m/s increase in velocity of sound in air, the temperature should rise by

\frac{1}{0.6} {}^{o}C

.
For 12 m/s increase in velocity of sound in air, the temperature should rise by =

\frac{12}{0.6} {}^{o}C = 20 {}^{o}C

Thus, the temperature at which the velocity of air becomes 344 m/s = ToC = 20oC

-

0oC = 20oC

b. Let the distance between Nita and the source of lightning be x.
Speed of sound in air = 340 m/s
Time taken by sound to reach Nita = 4 s
x = 340

\times 4

= 1360 m

c.

1. Time to hear first echo = 4 s
Distance travelled by sound in these 4 s = 360

\times 2

m
Speed of sound =

\frac{360 \times 2}{4} = 180 m / s

2. Time taken to hear second echo = 4 s + 2 s = 6 s
Distance travelled in these 6 s = 2x
Speed of sound in air = 180 m/s
Therefore, 2x = 180

\times 6

x = 540 m
Distance between the walls = 540 + 360 = 900 m

d. The velocity of sound in gas is related to density of gas as

v \propto \frac{1}{\sqrt{ρ}}

.....(i)
and the velocity of sound in gas is related to temperature of gas as

v \propto \sqrt{T}

.....(ii)
Combining (i) and (ii), we get

v \propto \frac{\sqrt{T}}{\sqrt{ρ}}

Now, the two bottles given are identical i.e. the volumes of gases are same. Let the volume of the bottle be V. Let M1 and M2 be the masses of the gases in bottles A and B, respectively and v1 and v2be the velocity of the sound in the two bottles, respectively. Since, the temperature of gas in both the bottles are same, let that common temperature be T. Therefore,

\frac{v_{1}}{v_{2}} = \frac{\sqrt{\frac{M_{2}}{V}} \sqrt{T}}{\sqrt{\frac{M_{1}}{V}} \sqrt{T}} = \frac{\sqrt{48}}{\sqrt{12}} = \sqrt{2} \Rightarrow v_{1} = 2 v_{2}

Thus, the sound travels 2 times faster in bottle B as compared to bottle A.

e. Mass of the gas in bottle A = 10 g
Mass of gas in bottle B = 40 g
The velocity of sound in gas is related to density of gas as

v \propto \frac{1}{\sqrt{ρ}}

.....(i)
and the velocity of sound in gas is related to temperature of gas as

v \propto \sqrt{T}

.....(ii)
Combining (i) and (ii), we get

v \propto \frac{\sqrt{T}}{\sqrt{ρ}}

The bottle are identical, this means the volumes of the gases are equal. Let the volume of the bottle be V. Let M1 and M2 be the masses of gases in bottles A and B, respectively and v1 and v2 be the velocity of the sound in the two bottles, respectively. Also, let T1 and T2 be their respective temperatures. Therefore,