Understanding the Doppler Equation: Effects of Moving Sources and Detectors

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When both the source and detector are moving in the same direction at the same speed, there is no relative speed between them, resulting in no Doppler effect. Consequently, the sign in the Doppler equation becomes irrelevant in this scenario. The equation indicates that if the speeds are equal, the observed frequency remains unchanged. Therefore, no frequency shift occurs when the source and detector are synchronized in speed and direction. Understanding this principle is crucial for accurately applying the Doppler equation in practical situations.
Ishu
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hi what will happen to teh sign on doppler's general equation when both source and dectector are moving in the same direction at the same speed.? I didnt find this on my books.? Does anyone knows?

dopler equation

F' = f (V +/- Vd)
------------
(V +/- Vs)


?
 
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Does anybody know about this?
 
if both source and dectector are moving in the same direction at the same speed, is there a relative speed between them?

then, does it matter what the sign is?
 
yeah both of them are at same direction at same speed 160 km
 
There won't be any doppler effect when both the observer and source is moving at the same rate of speed.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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