A police car sounding a siren with a frequency of 1580 Hz is t...

Question

A police car sounding a siren with a frequency of 1580 Hz is traveling at 120.0 km/h. What siren frequencies are heard in a car traveling at 90.0 km/h in the opposite direction before and after passing the police car?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice prom together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A train is moving at 115 kilometers per hour blows its whistle with a frequency of 1320 Hertz. A physics student attempts to measure the apparent frequency of the train's whistle on a bicycle moving at 72.0 kilometers per hour in the opposite direction towards the train. What is the measured frequency of the train's whistle before and after she passes the train assume that the speed of sound in air is 343 m per second. So that's our end goal. Our end goals were ultimately trying to figure out what the measured frequency of the train's whistle is. Before and after this physics student passes the train. So we're solving for two separate answers ultimately. Awesome. So now that we have an idea of what we're solving for, let's look at our multiple choice answers to see what our final answer pair might be. And let us quickly note that we're given an answer for before passing the train, which is denoted as FB. And we're also given an answer for after passing the train, which is denoted as FA or FB is the frequency before and FA is the frequency after. And all of our frequency values are in units of Hertz. So A is 542,140 B is 532,130 C is 2237 125 D is 440,066. OK. So first off, let us recall and use the equation to help us solve for the frequency before she passes the train. So let's say FB, we'll denote our frequency before is equal to F where F is the frequency of the sound source, which in this case, it's the sound source for our whistle, which is 1320 Hertz. And we need to multiply this value by VS lo vo and this will be all divided by VS minus VSC where VS in this case is the speed of sound in the air and vo is the speed of the observer on the bicycle and VSC is the speed of the source which in this particular case, the speed of the source is the train. So the train speed. And once again, FB is the apparent frequency on the bicycle before passing the train. And let us note that the bicycle is moving towards the train. Ok. So now at this point, we can plug in all of our known variables to solve for FB. So FB is equal to 1320 Hertz because that is the frequency of our sound source multiplied by the speed of sound and air, which is 343 m per second plus. Now, the speed of our p our physics student on the bicycle is 72.0 kilometers per hour. But to keep things simplistic, let's just call this 72 so 72 kilometers per hour. So now we have to use dimensional analysis to convert kilometers per hour to meters per second. So let us note that in one hour, there is 3600 seconds and let us note that there's 1000 m in one kilometer. Awesome. And don't forget your bracket. And this is all divided by another bracket which was our speed of sound. And ar again, which is 343 m per second minus which the speed of our train is 115 kilometers per hour. And once again, using dimensional analysis in one hour, there is 3600 seconds multiplied by 1000 m in one kilometer. Awesome. So let me plug that into our calculator. And we round to the nearest F number, we will find that FB is equal to 1540 Hertz. And that's it. We've solved for our first answer. We are on a roll. So now all we need to do is solve for FA and then we'll be done. So, moving right along. So similarly, we must recall and use the equation to softer the frequency after she passes the tree. So fa so this is our frequency after is equal to f multiplied by, in this case, it's gonna change a little bit. So it's gonna be vs minus vo divided by VS plus VSC. So like before we can now plug in all of our known variables to solve for FA. And let's quickly note once again that fa is the apparent frequency on the bicycle after passing the train as the bicycle is moving away from the train. Ok. So with that in mind, let's solve for FA so plugging in our known variables. So we know that FA is equal to 1320 Hertz. Awesome. And we need to multiply this by 343 m per second multiplied or sorry. So 343 m per second minus 72 kilometers per hour. Multiplied by one hour is approximately or I should say one hour is equivalent to 3600 seconds multiplied by 1000 m in one kilometer. Awesome. Divided by. So we're using dimensional analysis again, 343 m per second. Last 115 kilometers per hour, multiplied by one hour, divided by 3600 seconds, multiplied by 1000 m in one kilometer. Awesome. So like before when we plug this indoor calculator and we round to the nearest hole number, we will get 1140 Hertz and that's it we've solved for this problem. Ho Right. We did it. So looking at our multiple choice answers, the correct answer has to be the letter A, which letter A states that before passing the train FB is equal to 1540 Hertz. And after passing the train fa is equal to 1140 Hertz. And that's it. We've solved this problem. Hooray. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

A police car sounding a siren with a frequency of 1580 Hz is traveling at 120.0 km/h. What siren frequencies are heard in a car traveling at 90.0 km/h in the opposite direction before and after passing the police car?

Key Concepts

Doppler Effect

Relative Velocity

Sound Wave Propagation

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