A submarine with a cruising speed of 15.0 m/s uses a sonar (Sound Navigation and Ranging) device operating at 15000 Hz. The speed of sound in sea water at the submarine's depth is 1551 m/s....

 

A submarine with a cruising speed of 15.0 m/s uses a sonar (Sound Navigation and Ranging) device operating at 15000 Hz. The speed of sound in sea water at the submarine's depth is 1551 m/s. Calculate the frequency of the echo a sonar operator would receive from a torpedo coming directly towards the submarine at a speed of 30.0 m/s.

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quirozd | High School Teacher | (Level 3) Adjunct Educator

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Answer: 15896-16000 Hz (depending on rounding requested by your instructor) 

Background: SONAR is a bit more complicated, so we'll assume the easy case based on the information given. This question involves the Doppler effect with a reflection. In most problems, either the source or the observer is moving. In this case, the source is moving, sends a signal, which is perfectly reflected by a moving object, and observed with a Doppler shift. Typically, a beat (interference) is produced and the relative velocity of an object can be ascertained. 

Equations needed: 

`lambda_(apparent)=(c+-v_s)/f_s` Where f is the source frequency, v_s is the speed of the source and c is the speed of sound in the medium.

`f_(apparent)=(c+-v_o)/lambda_(apparent)`  where v_o is the speed of the observer. We choose + for observer moving towards, and - for moving away.

Putting these together gets us:

`f_(apparent)=(c+-v_o)/(c+-v_s)f_s`

Approach: We will apply the equation above twice. Once for the original signal, and again for the bounced signal. In the first part, the source is the submarine and the observer is the torpedo. In the second part, the roles reverse, using the observed frequency (perfectly bouncing off of the torpedo) as the source frequency. The final result will be the answer. We will also see if there is a faster/general way of doing this.

Plan and Calculate: 1st part: Submarine is source. Torpedo is observer.

`f_(apparent)=(c+v_o)/(c-v_s)f_s` Here, we chose + for observer and - for source because they are moving towards each other.

`f_(apparent1)=(1551 m/s+30 m/s)/(1551 m/s-15 m/s)15000 Hz=15439...Hz`

I would leave this number in my calculator to continue. This is the frequency of the sound as it hits the torpedo. Now for the reflection where f_apparent becomes f_s and the v_o is now the submarine.

`f_(apparent2)=(c+v_o)/(c-v_s)f_(apparent1)`  again, we chose the appropriate signs.

`f_(apparent2)=(1551 m/s+15 m/s)/(1551 m/s-30 m/s)15439...Hz=15896...Hz`

`~~15900Hz`

Evaluate: A sonar operator would observe a beat frequency of 900 Hz.

If we go back and look at the math, we see that we can create a general equation for this, so we can do it in one step.

`f_(text(received))=((c+v_text(target))(c+v_text(sender)))/((c-v_text(target))(c-v_text(sender)))`

Answer: 15900 Hz

Sources:

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