Sonar
Dear students, in this article we are going to deal with the interesting application of reflection of sound, which helps submarines and ships to locate heavy objects, enemy ships under ocean water.
SONAR works on principle of reflection of sound. SONAR is short form of “SOund Navigation And Ranging”. It uses the formula for velocity of sound to measure the depth of ocean or to find the submarines in the water. Refer the following diagram to understand the concept of working of SONAR in details.
SONAR is fixed to ship/submarine so the depth of ocean can be easily determined. It consist of two ports namely Transmitter and Receiver (Detector).
Transmitter continuously produces sound waves of sufficiently higher frequencies (e.g. ultrasonic waves) and sends it in to ocean water. These sound waves travel through the water with velocity ‘v’. Now suppose that there is a rock stone present at depth ‘d’ of the ocean water. The waves emitted from the transmitter travels distance ‘d’ and reaches to depth of ocean and gets reflected back with the same angle and velocity towards the ship.
Receiver receives the reflected waves from the depth of ocean and reads the time between transmission of waves and receiving after reflection. By knowing the velocity of sound waves and the time interval between transmission and reception, one can find the depth ocean.
Applications of SONAR:
- It is uses to find the depth of ocean surface.
- It can detect the present of rock, hills.
- It can detect hidden submarines which helps the neavy forces to find location of enemy forces to prevent any hidden attack.
Let’s solve the numeric given below to understand the more about the SONAR.
Ex: 1) Ultrasonic wave are sent by transmitter of SONAR travels with velocity of 1450 m/s in ocean water strikes on underwater hill and received by receiver in 6 seconds after transmission. Find the distance of underwater hill from the surface of water.
Ans: vwater =1450 m/s ,
t = 6 s
We have, v=d/t
∴d = 1450×6
∴d = 8700 m
Now this is the total distance travelled by when during interval of transmission and reception. Then the distance between the surface of water and underwater hill will be half of this distance,
∴ d’ = 8700/2 = 4350 m = 4.35 km