Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
The frequency produced by the siren is 631.12 Hz
Doppler effect
The variation in frequency when a source of sound moves relative to an observer is determined by the doppler effect.
Frequency of observer
So, the frequency of the observer f' = (v ± v')f/(v ± v") where
f' = 590 Hz
f = frequency of source or siren ,
v = speed of sound = 330 m/s,
v' = speed of observer = 0 m/s (since you are stationary) and
v" = speed of source = 23 m/s
Since the source moves away from the detector, the sign in the denominator is positive and v' = 0 m/s
So, f' = (v + 0)f/(v + v")
f' = vf/(v + v")
Since, we require the frequency of the source, make f subject of the formula, we have
Frequency of siren
f = (v + v")f'/v
Substituting the values of the variables into the equation, we have
f = (v + v")f'/v
f' = (330 m/s + 23 m/s) × 590 Hz/330 m/s
f' = 353 m/s × 590 Hz/330 m/s
f' = 208270 m/sHz/330 m/s
f' = 631.12 Hz
The frequency produced by the siren is 631.12 Hz
Learn more about doppler effect here:
https://brainly.com/question/2169203
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.