Finding the Distance Between Sound and Wave Maxima at a Train Station

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timetraveller123
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Homework Statement


Two trains approach the train station from opposite sides each moving at an initialvelocity of 120 km/h with respect to the station and slowing down to a halt in 20seconds. If both trains are whistling at 5000 Hz, find the distance between the zeroand first order maxima as a function of time.
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so this is the question:
when they say zero and first order maxima they mean it only lies on the x-axis right meaning the zero order maxima is at the centre of the two trains and the first order maxima is at distance d away where the extra length traveled is exactly a wavelength so i did the question just for the part when the trains are not decelerating ie t=0 and my solution does not match the given one please help or is the first order maxima in the y direction

Homework Equations


f = fo [c/c-v] --- doppler effect for train approaching
c= 331m/s
λ = c/f
v = 120/3.6 m/s
the given solution is at t =0 the distance between the zero and first order is 893/60000 m

The Attempt at a Solution


the first order is offset to one side by distance d so
2d is the extra length traveled the sound wave coming from the other side hence
2d=λ
2d = c/f
2d = c(c-v)/(2c fo)
plugging in gets me 893/30000 which is twice of correct answer please help
 
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vishnu 73 said:
2d is the extra length traveled the sound wave coming from the other side
True, but doesn't the one from the opposite side travel over a little less distance ?
 
BvU said:
True, but doesn't the one from the opposite side travel over a little less distance ?
i am sorry i don't get what you mean could please explain your point thanks
 
Make a drawing and you'll see it more easily. Constructive interference occurs where one wave travels ##{1\over 2}\lambda ## further and the other ##{1\over 2}\lambda ## less.
 
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BvU said:
Make a drawing and you'll see it more easily. Constructive interference occurs where one wave travels ##{1\over 2}\lambda ## further and the other ##{1\over 2}\lambda ## less.
oh i see it now thanks so much !
 
no wait sir i did what you said and all of a sudden i am confused again as i once again get d =λ/2 please enlighten me as if one wave were to travel 1/2λ lesser it implies that that is d
 
BvU said:
Make a drawing and you'll see it more easily. Constructive interference occurs where one wave travels ##{1\over 2}\lambda## further and the other ##{1\over 2}\lambda## less.
I made the drawing and find I have to correct myself: not ##{1\over 2}\lambda## but ##{1\over 4}\lambda## ... o:)

Can you see why ?

(It's astounding to see how much more instructive an animated picture is wrt one on paper ! How wonderful all these newfangled technologies o0) )
 
thank you sir i figured out my mistake i was all along using cos graph for both side only to realize that when the two cos graphs collide they will super pose to zero at every point i should have used cos graph on one side and -cos graph on the other side and then your 1/4λ thing works thanks for your troubles really appreciate it !