Solution-Class 9-Science-Chapter-12-Study of Sound-Maharashtra Board

a. Sound does not travel through vacuum.

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

c. The incidence of thunderstormin 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, SONARtechnology is used.

The persistence of sound due to repeated reflection of sound in a big hall or an auditorium is called reverberation.

The roof of a movie theatre and a conference hall is curved to avoid undesirable reverberation. Otherwise, pitch of music will not be heard distinctly.

In an empty house, there is no sound absorbers present such as furniture, curtains etc. In a closed and empty house, there is absorption of sound only by the floor, walls and the ceiling. Hence, the intensity of reverberation is higher.

To hear distinct echo, the minimum distance between the source of sound and obstacle should be 17.2 m (assuming room temperature as 22°C). The classrooms are designed in such a way that, this condition is not satisfied in our classroom as the ceiling is not so high and the distance between the walls are less than 17.2 m. Hence, we cannot hear echo produced in our classroom

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 sensation of sound lasts in our brain for about 0.1 s. The distance covered by sound in this time = speed of sound x time = 344 m/s x 0.1 s = 34.4 m as the speed of sound in air at 22°C is 344 m/ s.

Total distance (source to obstacle and back),

34.4 m = 2 x distance between the source and obstacle.

∴ Distance between the source and obstacle = 34.4m/2 = 17.2 m

This gives the minimum distance between the source and obstacle for hearing distinct echo in the present case.

• An echo can be heard multiple times in the golghumat located in Vijaypur, Karnataka because of multiple or continuous reflections on its surfaces.
• 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.

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

The sound absorbing materials are used on the roofs and walls of auditoriums, concert halls, theatres, etc. to avoid production of echo.

In this case, v at t°C = (v at 0°C )+ (0.6 t) m/s

Hence, by the data given,

344 = 332 + 0.6 t

∴ t = \(\frac{344-332}{0.6}\) = 20°C

This will be the required temperature

Given : t= 4 s, v (sound) = 340 m/s

The speed of light in air is about 3 x 10⁸ m/s.

This is far greater than v (sound) in air.

Hence, the required distance,

S = vt =340 x 4 =1360 m

Given : Time to hear first echo t₁= 4s, Time taken to hear second echo t₂= 4 s + 2 s = 6 s

Time required for first echo to be heard 4s = 2 x time required to reach the first wall (t₁)

t₁ =4/2 s = 2s

Time required for second echo to be heard 6s = 2 x time required to reach the second wall (t₂)

t₂ =6/2 s = 3s

Velocity of sound in air,

V = s/t =  m/s = 330 m/s

Distance between the two walls,

= s₁ + s₂

= vt₁ + vt₂

= v(t₁ + t₂)

= 330(2+3)

=330 x 5 = 1650 m

The velocity of sound in gas is related to density of gas as
v ∝ \(\sqrt{\frac{1}{ρ}}\)   .....(i)
and the velocity of sound in gas is related to temperature of gas as
v ∝ \(\sqrt{T}\)   .....(ii)
from eq. (i) and (ii), we get
v ∝ \(\sqrt{\frac{T}{ρ}}\) ∝ \(\sqrt{\frac{VT}{M}}\)     ……..(as ρ = M/V)
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 v₂ be 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,
v₁/v₂ = \(\sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{42}{12}}\)  =  2

 v₁ = 2v2
Thus, the sound travels 2 times faster in bottle A.

For ref. see above solved example.

\(v ∝ \sqrt{\frac{VT}{M}}\)

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 A and B bottles, respectively. Also, let T₁ and T₂ be their respective temperatures. Therefore,

 \(v_1 ∝\sqrt{\frac{VT_1}{M_1}}\)  & \(v_2 ∝\sqrt{\frac{VT_2}{M_2}}\)

since the speed of the sound is same 

∴  \(\sqrt{\frac{VT_1}{M_1}}\)  = \(\sqrt{\frac{VT_2}{M_2}}\)

∴ T₂  = \(\frac{M_2T_1}{M_1} = \frac{40T_1}{10}\)

∴ T₂ =  4T₁

∴ Temperature of B is 4 times the temperature of A.