A SONAR system fixed in a submarine operates at a frequency of 40.0kHz. An enemy submarine moves towards the SONAR with a speed of 360km/h. Taking the speed of sound in water as 1450m/s, then frequency of the reflected sound is
A. 46kHz
B. 51kHz
C. 55kHz
D. 60kHz
Answer
582.3k+ views
Hint: You could first read the question well and hence write down the given values from the question. Then you could consider the two cases: one before reflection. Apply the concept of doppler effect with the help of the formula related to it and hence solve the problem.
Complete step-by-step solution:
In the question, we are given a SONAR system that is fixed in a submarine that is operating at a frequency of 40kHz. Now we have an enemy submarine that is moving towards this SONAR at a speed of 350km/h. We are supposed to find the frequency of the reflected sound by considering the speed of sound in water to be 1450m/s.
So, we are given a problem based on the doppler effect.
Here the source is said to be at rest and the observer which would be the enemy submarine is moving towards it.
SONAR system’s frequency,
$f=40kHz=40\times {{10}^{3}}Hz$
Velocity of the observer is given to be,
${{v}_{o}}=360km/h=360\times \dfrac{5}{18}m/s=100m/s$
Speed sound in water
v=1450m/s
Now we have the apparent frequency of the sound wave to be,
$f'=\left( \dfrac{v+{{v}_{o}}}{v} \right)f=\dfrac{1450+100}{1450}\times 4\times {{10}^{4}}$……………………………….. (1)
Now this is perceived by the enemy submarine and would be reflected. Here the source is moving with velocity,
${{v}_{s}}=100m/s$
Now we have the frequency of the reflected wave given by,
${{f}_{reflected}}=\left( \dfrac{v}{v-{{v}_{s}}} \right)f'$
$\Rightarrow {{f}_{reflected}}=\dfrac{1450}{1450-100}\times \dfrac{1450+100}{1450}\times 4\times {{10}^{4}}$
$\therefore {{f}_{reflected}}=45.92\times {{10}^{3}}=45.92kHz$
Therefore, we found the velocity of the reflected sound to be approximately 46kHz. Hence, option A is true.
Note: In the question, we actually have two cases to deal with. For the first case the source is at rest and the receiver is moving and for the second case, the source is moving with the receiver at rest. The second case is for the case of the reflected sound.
Complete step-by-step solution:
In the question, we are given a SONAR system that is fixed in a submarine that is operating at a frequency of 40kHz. Now we have an enemy submarine that is moving towards this SONAR at a speed of 350km/h. We are supposed to find the frequency of the reflected sound by considering the speed of sound in water to be 1450m/s.
So, we are given a problem based on the doppler effect.
Here the source is said to be at rest and the observer which would be the enemy submarine is moving towards it.
SONAR system’s frequency,
$f=40kHz=40\times {{10}^{3}}Hz$
Velocity of the observer is given to be,
${{v}_{o}}=360km/h=360\times \dfrac{5}{18}m/s=100m/s$
Speed sound in water
v=1450m/s
Now we have the apparent frequency of the sound wave to be,
$f'=\left( \dfrac{v+{{v}_{o}}}{v} \right)f=\dfrac{1450+100}{1450}\times 4\times {{10}^{4}}$……………………………….. (1)
Now this is perceived by the enemy submarine and would be reflected. Here the source is moving with velocity,
${{v}_{s}}=100m/s$
Now we have the frequency of the reflected wave given by,
${{f}_{reflected}}=\left( \dfrac{v}{v-{{v}_{s}}} \right)f'$
$\Rightarrow {{f}_{reflected}}=\dfrac{1450}{1450-100}\times \dfrac{1450+100}{1450}\times 4\times {{10}^{4}}$
$\therefore {{f}_{reflected}}=45.92\times {{10}^{3}}=45.92kHz$
Therefore, we found the velocity of the reflected sound to be approximately 46kHz. Hence, option A is true.
Note: In the question, we actually have two cases to deal with. For the first case the source is at rest and the receiver is moving and for the second case, the source is moving with the receiver at rest. The second case is for the case of the reflected sound.
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