Chapter : 5. Sound
Examples on Sonar
Example. 11
The ultrasonic waves take 4 seconds to travel from the ship to the bottom of the sea and back to the ship. What is the depth of the sea ? (Speed of sound in water = 1500 m/s.)
Solution. The time taken by the ultrasonic sound waves to travel from the ship to the sea-bed and back to the ship is 4 seconds. So, the time taken by the ultrasonic sound to travel from the ship to sea-bed will be half of this time, which is seconds. This means that the sound takes 2 seconds to travel from the ship to the bottom of the sea
Now,
So, 1500 = Distance/2
And, Distance = 1500 × 2m = 3000m
Example. 12
A submarine emits a sonar pulse which returns from the underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 ms–1, how far away is the cliff ?
Solution. Given : Speed of sonar pulse, V = 1531 ms–1, Time interval of return journey of the pulse, t = 1.02s
Let the distance of the underwater cliff be S.
For distance S of the cliff, the pulse travels a total distance of 2S in return journey.
From relation, distance = speed × time
2S = vt
We have, S = vt/2
S = 780.8 m
The ultrasonic waves take 4 seconds to travel from the ship to the bottom of the sea and back to the ship. What is the depth of the sea ? (Speed of sound in water = 1500 m/s.)
Solution. The time taken by the ultrasonic sound waves to travel from the ship to the sea-bed and back to the ship is 4 seconds. So, the time taken by the ultrasonic sound to travel from the ship to sea-bed will be half of this time, which is seconds. This means that the sound takes 2 seconds to travel from the ship to the bottom of the sea
Now,
So, 1500 = Distance/2
And, Distance = 1500 × 2m = 3000m
Example. 12
A submarine emits a sonar pulse which returns from the underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 ms–1, how far away is the cliff ?
Solution. Given : Speed of sonar pulse, V = 1531 ms–1, Time interval of return journey of the pulse, t = 1.02s
Let the distance of the underwater cliff be S.
For distance S of the cliff, the pulse travels a total distance of 2S in return journey.
From relation, distance = speed × time
2S = vt
We have, S = vt/2
S = 780.8 m
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