a)if we place the porous surface at the back of the hollow tubes in the experimental setup of the reflection of sound, then (i)sound will be heard with greater intensity than the incident sound. (ii)sound will be heard with lesser intensity than the incident sound. (iii)reflection of sound does not take place. (iv)reflected sound will remain the same as incident sound. (b)explain why.
Answers
Reflection of Sound
Sound bounces off a solid or a liquid like a rubber ball bounces off a wall. Like light, sound gets reflected at the surface of a solid or liquid and follows the same laws of reflection as you have studied in earlier classes. The directions in which the sound is incident and is reflected make equal angles with the normal to the reflecting surface at the point of incidence, and the three are in the same plane.
Activity
• Take two identical pipes, as shown in Fig. 12.11. You can make the pipes using chart paper. The length of the pipes should be sufficiently long as shown.
• Arrange them on a table near a wall.
• Keep a clock near the open end of one of the pipes and try to hear the sound of the clock through the other pipe.
• Adjust the position of the pipes so that you can best hear the sound of the clock.
• Now, measure the angles of incidence and reflection and see the relationship between the angles.
• Lift the pipe on the right vertically to a small height and observe what happens.
If we shout or clap near a suitable reflecting object such as a tall building or a mountain, we will hear the same sound again a little later. This sound which we hear is called an echo. The sensation of sound persists in our brain for about 0.1 s. To hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1s. If we take the speed of sound to be 344 m/s at a given temperature, say at 22 ºC in air, the sound must go to the obstacle and reach back the ear of the listener on reflection after 0.1s. Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least (344 m/s) × 0.1 s = 34.4 m.
Sonar
The acronym SONAR stands for SOund Navigation And Ranging. Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. How does the sonar work? Sonar consists of a transmitter and a detector and is installed in a boat or a ship, as shown in Fig.
The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the
seabed, get reflected back and are sensed by the detector. The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound. Let the time interval between transmission and reception of ultrasound signal be t and the speed of sound through seawater be v. The total distance, 2d travelled by the ultrasound is then, 2d = v × t.
The above method is called echo-ranging. The sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunken ship etc.
Example
A ship sends out ultrasound that returns from the seabed and is detected after 3.42 s. If the speed of ultrasound through seawater is
1531 m/s, what is the distance of the
seabed from the ship?
Solution: Given, Time between transmission and
detection, t = 3.42 s.Speed of ultrasound in sea water, v = 1531 m/s
Distance travelled by the ultrasound = 2 × depth of the sea = 2d
where d is the depth of the sea.
2d = speed of sound × time = 1531 m/s × 3.42 s = 5236 m
d = 5236 m/2 = 2618 m.
Thus, the distance of the seabed from the ship is 2618 m or 2.62 km.