I know how to solve to get the speed of sound but are they moving forward (+) or backward (-) and what's is the observer and source. Is the source the police siren?vela said:What specifically is confusing you?
Do I use the formula I gave or the one from the picture?vela said:What specifically is confusing you?
vela said:The observer is what detects the sound. The source is what produces the sound the observer hears.
Consider the two motions separately. What effect does the motion of the police car have on the frequency detected, and what effect does your movement have on the frequency detected?
348m/s is the speed of soundThePhysicsXV said:So is V = 988(348+7/ 348+8)
Or V=988(348-7/348-23)?
I have no idea how to answer your question.vela said:Instead of plugging numbers in randomly, please answer the questions.
How would you "solve to get the speed of sound"? The speed of sound is a given.ThePhysicsXV said:I know how to solve to get the speed of sound but are they moving forward (+) or backward (-) and what's is the observer and source. Is the source the police siren?
The source and observer are moving on the same direction since the police car coming toward you, and you getting away.haruspex said:How would you "solve to get the speed of sound"? The speed of sound is a given.
As vela says, the source is the siren, you are the observer.
Are the source and observer moving in the same direction or in opposite directions?
As vela suggests, try it in two parts. First, what frequency did you hear before starting to run? Can you quote an equation for the Doppler effect?
Ok, but what about the Doppler effect? So far, I see no evidence that you've even heard of it.ThePhysicsXV said:The source and observer are moving on the same direction since the police car coming toward you, and you getting away.
V sound in air = 331.4 + 0.6T
T is temperature so temp=28
And you will get 348m/s
haruspex said:Ok, but what about the Doppler effect? So far, I see no evidence that you've even heard of it.
Ok, so apply that.ThePhysicsXV said:Doppler Effect for Moving Source and Observer
ƒ'=ƒ((1±u₀/v)/(1±u(sound)/v))
Actually they both moving at different speeds. Police at 23m/s and "you" at 7m/sharuspex said:Ok, so apply that to the case where the car is approaching you but you are standing still.
Yes, I replied a little too quickly, then edited it.ThePhysicsXV said:Actually they both moving at different speeds. Police at 23m/s and "you" at 7m/s
haruspex said:Yes, I replied a little too quickly, then edited it.
I anticipated that you would quote a formula for only one movement, so I was trying to get you to solve it in the two stages, as vela had suggested. When I realized your equation allowed for both moving I edited my post, but you were too quick for me.
So, apply the equation.
haruspex said:Yes, I replied a little too quickly, then edited it.
I anticipated that you would quote a formula for only one movement, so I was trying to get you to solve it in the two stages, as vela had suggested. When I realized your equation allowed for both moving I edited my post, but you were too quick for me.
So, apply the equation.
f= f(v-o/v-s)haruspex said:I did have some concerns about the equation you posted previously, but I wanted to see how you applied it. The equations in the image you have attached look fine, except that you should ignore the references to "Stationary" observer. That makes no sense given the rest of the text. It's probably a cut-and-paste error from some preceding cases.
So, which of the two cases in the image do you think applies here? What do you get when you apply it?
I get the same. Round that to the appropriate number of digits.ThePhysicsXV said:f= f(v-o/v-s)
I got 1036.64Hz
Okay, I guess that's the correct answer. But Thank you sir for your helpharuspex said:I get the same. Round that to the appropriate number of digits.